\displaystyle \textbf{Evaluate the following integrals as limit of sums:}

\displaystyle \textbf{Question 1: }~\int_{0}^{3}(x+4)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=3,\ f(x)=x+4,\ h=\frac{3-0}{n}=\frac{3}{n}
\displaystyle \text{Therefore, } I=\int_0^3 (x+4)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0+4)+(h+4)+(2h+4)+\cdots+((n-1)h+4)]
\displaystyle =\lim_{n\to\infty}h\,[4n+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[4n+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{3}{n}\left[4n+\frac{3}{n}\cdot\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\left[12+\frac{9}{2}\left(1-\frac{1}{n}\right)\right]
\displaystyle =12+\frac{9}{2}=\frac{33}{2}

\displaystyle \textbf{Question 2: }~\int_{0}^{2}(x+3)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=x+3,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 (x+3)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0+3)+(h+3)+(2h+3)+\cdots+((n-1)h+3)]
\displaystyle =\lim_{n\to\infty}h\,[3n+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[3n+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[3n+\frac{2}{n}\cdot\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\left[6+2\left(1-\frac{1}{n}\right)\right]
\displaystyle =8

\displaystyle \textbf{Question 3: }~\int_{1}^{3}(3x-2)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=1,\ b=3,\ f(x)=3x-2,\ h=\frac{3-1}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_1^3 (3x-2)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(3\cdot1-2)+(3(1+h)-2)+(3(1+2h)-2)+\cdots+(3(1+(n-1)h)-2)]
\displaystyle =\lim_{n\to\infty}h\,[n+3h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[n+3h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[n+3\cdot\frac{2}{n}\cdot\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\left[2+6\left(1-\frac{1}{n}\right)\right]
\displaystyle =8

\displaystyle \textbf{Question 4: }~\int_{-1}^{1}(x+3)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=-1,\ b=1,\ f(x)=x+3,\ h=\frac{1-(-1)}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_{-1}^1 (x+3)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(-1)+f(-1+h)+f(-1+2h)+\cdots+f(-1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(-1+3)+(-1+h+3)+(-1+2h+3)+\cdots+(-1+(n-1)h+3)]
\displaystyle =\lim_{n\to\infty}h\,[2n+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[2n+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[2n+\frac{2}{n}\cdot\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\left[4+2\left(1-\frac{1}{n}\right)\right]
\displaystyle =6

\displaystyle \textbf{Question 5: }~\int_{0}^{1}(x+1)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=5,\ f(x)=x+1,\ h=\frac{5-0}{n}=\frac{5}{n}
\displaystyle \text{Therefore, } I=\int_0^5 (x+1)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0+1)+(h+1)+(2h+1)+\cdots+((n-1)h+1)]
\displaystyle =\lim_{n\to\infty}h\,[n+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[n+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{5}{n}\left[n+\frac{5}{n}\cdot\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}5\left[\frac{7}{2}-\frac{5}{2n}\right]
\displaystyle =\frac{35}{2}

\displaystyle \textbf{Question 6: }~\int_{1}^{3}(2x+3)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=1,\ b=3,\ f(x)=2x+3,\ h=\frac{3-1}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_1^3 (2x+3)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(2\cdot1+3)+(2(1+h)+3)+(2(1+2h)+3)+\cdots+(2(1+(n-1)h)+3)]
\displaystyle =\lim_{n\to\infty}h\,[5n+2h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[5n+2h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[5n+2\cdot\frac{2}{n}\cdot\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\left[10+4\left(1-\frac{1}{n}\right)\right]
\displaystyle =14

\displaystyle \textbf{Question 7: }~\int_{3}^{5}(2-x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=3,\ b=5,\ f(x)=2-x,\ h=\frac{5-3}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_3^5 (2-x)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(3)+f(3+h)+f(3+2h)+\cdots+f(3+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(2-3)+(2-(3+h))+(2-(3+2h))+\cdots+(2-(3+(n-1)h))]
\displaystyle =\lim_{n\to\infty}h\,[-n-h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[-n-h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[-n-\frac{2}{n}\cdot\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\left[-2-2\left(1-\frac{1}{n}\right)\right]
\displaystyle =-4

\displaystyle \textbf{Question 8: }~\int_{0}^{2}(x^{2}+1)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=x^2+1,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 (x^2+1)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0^2+1)+(h^2+1)+(2h)^2+1+\cdots+((n-1)^2h^2+1)]
\displaystyle =\lim_{n\to\infty}h\,[n+h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[n+h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[n+\frac{4}{n^2}\cdot\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}2\left[1+\frac{2}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle =2+\frac{8}{3}=\frac{14}{3}

\displaystyle \textbf{Question 9: }~\int_{1}^{2}x^{2}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=1,\ b=2,\ f(x)=x^2,\ h=\frac{2-1}{n}=\frac{1}{n}
\displaystyle \text{Therefore, } I=\int_1^2 x^2\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[1^2+(1+h)^2+(1+2h)^2+\cdots+(1+(n-1)h)^2]
\displaystyle =\lim_{n\to\infty}h\,[n+2h(1+2+3+\cdots+(n-1))+h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[n+2h\frac{n(n-1)}{2}+h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{1}{n}\left[n+\frac{n(n-1)}{n}+\frac{(n-1)(2n-1)}{6n}\right]
\displaystyle =\lim_{n\to\infty}\left[1+\left(1-\frac{1}{n}\right)+\frac{1}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle =2+\frac{1}{3}=\frac{7}{3}

\displaystyle \textbf{Question 10: }~\int_{2}^{3}(2x^{2}+1)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=2,\ b=3,\ f(x)=2x^2+1,\ h=\frac{3-2}{n}=\frac{1}{n}
\displaystyle \text{Therefore, } I=\int_2^3 (2x^2+1)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(2)+f(2+h)+f(2+2h)+\cdots+f(2+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[2(2)^2+1+2(2+h)^2+1+2(2+2h)^2+1+\cdots+2(2+(n-1)h)^2+1]
\displaystyle =\lim_{n\to\infty}h\,[9n+8h(1+2+3+\cdots+(n-1))+2h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[9n+8h\frac{n(n-1)}{2}+2h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{1}{n}\left[9n+4(n-1)+\frac{(n-1)(2n-1)}{3n}\right]
\displaystyle =\lim_{n\to\infty}\left[13+\frac{1}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)-\frac{4}{n}\right]
\displaystyle =13+\frac{2}{3}=\frac{41}{3}

\displaystyle \textbf{Question 11: }~\int_{1}^{2}(x^{2}-1)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=1,\ b=2,\ f(x)=x^2-1,\ h=\frac{2-1}{n}=\frac{1}{n}
\displaystyle \text{Therefore, } I=\int_1^2 (x^2-1)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[0+(1+h)^2-1+(1+2h)^2-1+\cdots+(1+(n-1)h)^2-1]
\displaystyle =\lim_{n\to\infty}h\,[n-1+2h(1+2+3+\cdots+(n-1))+h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[n-1+2h\frac{n(n-1)}{2}+h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{1}{n}\left[n-1+(n-1)+\frac{(n-1)(2n-1)}{6n}\right]
\displaystyle =\lim_{n\to\infty}\left[1-\frac{1}{n}+\frac{1}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle =1+\frac{1}{3}=\frac{4}{3}

\displaystyle \textbf{Question 12: }~\int_{0}^{2}(x^{2}+4)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=x^2+4,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 (x^2+4)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0^2+4)+(h^2+4)+(2h)^2+4+\cdots+((n-1)^2h^2+4)]
\displaystyle =\lim_{n\to\infty}h\,[4n+h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[4n+h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[4n+\frac{4}{n^2}\cdot\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}2\left[4+\frac{2}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle =8+\frac{8}{3}=\frac{32}{3}

\displaystyle \textbf{Question 13: }~\int_{2}^{4}(x^{2}-x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=1,\ b=4,\ f(x)=x^2-x,\ h=\frac{4-1}{n}=\frac{3}{n}
\displaystyle \text{Therefore, } I=\int_1^4 (x^2-x)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[1^2-1+(1+h)^2-(1+h)+(1+2h)^2-(1+2h)+\cdots+(1+(n-1)h)^2-(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[h^2(1^2+2^2+3^2+\cdots+(n-1)^2)+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[h^2\frac{n(n-1)(2n-1)}{6}+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{3}{n}\left[\frac{9}{n^2}\cdot\frac{n(n-1)(2n-1)}{6}+\frac{3}{n}\cdot\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}3\left[\frac{3}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{3}{2}\left(1-\frac{1}{n}\right)\right]
\displaystyle =9+\frac{9}{2}=\frac{27}{2}

\displaystyle \textbf{Question 14: }~\int_{0}^{1}(3x^{2}+5x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=1,\ f(x)=3x^2+5x,\ h=\frac{1-0}{n}=\frac{1}{n}
\displaystyle \text{Therefore, } I=\int_0^1 (3x^2+5x)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0)+(3h^2+5h)+(3(2h)^2+5(2h))+\cdots+(3(n-1)^2h^2+5(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[5h(1+2+3+\cdots+(n-1))+3h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[5h\frac{n(n-1)}{2}+3h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{1}{n}\left[\frac{5(n-1)}{2}+\frac{(n-1)(2n-1)}{2n}\right]
\displaystyle =\lim_{n\to\infty}\left[\frac{5}{2}\left(1-\frac{1}{n}\right)+\frac{1}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle =\frac{5}{2}+1=\frac{7}{2}

\displaystyle \textbf{Question 15: }~\int_{0}^{2}e^{x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=e^x,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 e^x\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[e^0+e^h+e^{2h}+\cdots+e^{(n-1)h}]
\displaystyle =\lim_{n\to\infty}h\left[\frac{e^{nh}-1}{e^h-1}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[\frac{e^2-1}{e^{2/n}-1}\right]
\displaystyle =\lim_{n\to\infty}(e^2-1)\left[\frac{2/n}{e^{2/n}-1}\right]
\displaystyle =e^2-1

\displaystyle \textbf{Question 16: }~\int_{a}^{b}e^{x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=a,\ b=b,\ f(x)=e^x,\ h=\frac{b-a}{n}
\displaystyle \text{Therefore, } I=\int_a^b e^x\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[e^a+e^{a+h}+e^{a+2h}+\cdots+e^{a+(n-1)h}]
\displaystyle =\lim_{n\to\infty}h\left[e^a\{1+e^h+e^{2h}+\cdots+e^{(n-1)h}\}\right]
\displaystyle =\lim_{n\to\infty}h\left[e^a\frac{e^{nh}-1}{e^h-1}\right]
\displaystyle =\lim_{n\to\infty}\frac{b-a}{n}\left[e^a\frac{e^{b-a}-1}{e^{(b-a)/n}-1}\right]
\displaystyle =\lim_{n\to\infty}(e^b-e^a)\left[\frac{(b-a)/n}{e^{(b-a)/n}-1}\right]
\displaystyle =e^b-e^a

\displaystyle \textbf{Question 17: }~\int_{a}^{b}\cos x\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=a,\ b=b,\ f(x)=\cos x,\ h=\frac{b-a}{n}
\displaystyle \text{Therefore, } I=\int_a^b \cos x\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(a)+f(a+h)+\cdots+f(a+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[\cos a+\cos(a+h)+\cdots+\cos\{a+(n-1)h\}]
\displaystyle =\lim_{n\to\infty}h\left[\frac{\cos\!\left(a+\frac{(n-1)h}{2}\right)\sin\!\left(\frac{nh}{2}\right)}{\sin\!\left(\frac{h}{2}\right)}\right]
\displaystyle =\lim_{n\to\infty}\frac{h}{2\sin\!\left(\frac{h}{2}\right)}\,2\cos\!\left(a+\frac{b-a}{2}-\frac{h}{2}\right)\sin\!\left(\frac{b-a}{2}\right)
\displaystyle =\left[\lim_{n\to\infty}\frac{h}{2\sin\!\left(\frac{h}{2}\right)}\right]\left[\lim_{n\to\infty}2\cos\!\left(\frac{a+b}{2}-\frac{h}{2}\right)\sin\!\left(\frac{b-a}{2}\right)\right]
\displaystyle =2\cos\!\left(\frac{a+b}{2}\right)\sin\!\left(\frac{b-a}{2}\right)
\displaystyle =\sin b-\sin a

\displaystyle \textbf{Question 18: }~\int_{0}^{\pi/2}\sin x\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=\frac{\pi}{2},\ f(x)=\sin x,\ h=\frac{\frac{\pi}{2}-0}{n}=\frac{\pi}{2n}
\displaystyle \text{Therefore, } I=\int_0^{\frac{\pi}{2}} \sin x\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[\sin0+\sin h+\sin2h+\cdots+\sin((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\left[\frac{\sin\!\left(\frac{(n-1)h}{2}\right)\sin\!\left(\frac{nh}{2}\right)}{\sin\!\left(\frac{h}{2}\right)}\right]
\displaystyle =\lim_{n\to\infty}\frac{h}{2\sin\!\left(\frac{h}{2}\right)}\,2\sin\!\left(\frac{\pi}{4}-\frac{h}{2}\right)\sin\!\left(\frac{\pi}{4}\right)
\displaystyle =\left[\lim_{n\to\infty}\frac{h}{2\sin\!\left(\frac{h}{2}\right)}\right]\left[\lim_{n\to\infty}2\sin\!\left(\frac{\pi}{4}-\frac{h}{2}\right)\sin\!\left(\frac{\pi}{4}\right)\right]
\displaystyle =2\sin\!\left(\frac{\pi}{4}\right)\sin\!\left(\frac{\pi}{4}\right)
\displaystyle =2\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}=1

\displaystyle \textbf{Question 19: }~\int_{0}^{\pi/2}\cos x\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=\frac{\pi}{2},\ f(x)=\cos x,\ h=\frac{\frac{\pi}{2}-0}{n}=\frac{\pi}{2n}
\displaystyle \text{Therefore, } I=\int_0^{\frac{\pi}{2}} \cos x\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[\cos0+\cos h+\cos2h+\cdots+\cos((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\left[\frac{\cos\!\left(\frac{(n-1)h}{2}\right)\sin\!\left(\frac{nh}{2}\right)}{\sin\!\left(\frac{h}{2}\right)}\right]
\displaystyle =\lim_{n\to\infty}\frac{h}{2\sin\!\left(\frac{h}{2}\right)}\,2\cos\!\left(\frac{\pi}{4}-\frac{h}{2}\right)\sin\!\left(\frac{\pi}{4}\right)
\displaystyle =\left[\lim_{n\to\infty}\frac{h}{2\sin\!\left(\frac{h}{2}\right)}\right]\left[\lim_{n\to\infty}2\cos\!\left(\frac{\pi}{4}-\frac{h}{2}\right)\sin\!\left(\frac{\pi}{4}\right)\right]
\displaystyle =2\cos\!\left(\frac{\pi}{4}\right)\sin\!\left(\frac{\pi}{4}\right)
\displaystyle =2\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}=1

\displaystyle \textbf{Question 20: }~\int_{1}^{4}(3x^{2}+2x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=1,\ b=4,\ f(x)=3x^2+2x,\ h=\frac{4-1}{n}=\frac{3}{n}
\displaystyle \text{Therefore, } I=\int_1^4 (3x^2+2x)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(3\cdot1^2+2\cdot1)+(3(1+h)^2+2(1+h))+\cdots+(3(1+(n-1)h)^2+2(1+(n-1)h))]
\displaystyle =\lim_{n\to\infty}h\,[5n+6h(1+2+3+\cdots+(n-1))+3h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[5n+6h\frac{n(n-1)}{2}+3h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{3}{n}\left[5n+9(n-1)+\frac{9(n-1)(2n-1)}{2n}\right]
\displaystyle =\lim_{n\to\infty}3\left[17-\frac{12}{n}+\frac{9}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle =51+27=78

\displaystyle \textbf{Question 21: }~\int_{0}^{2}(3x^{2}-2)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=3x^2-2,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 (3x^2-2)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0-2)+(3h^2-2)+(3(2h)^2-2)+\cdots+(3(n-1)^2h^2-2)]
\displaystyle =\lim_{n\to\infty}h\,[-2n+3h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[-2n+3h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[-2n+\frac{2(n-1)(2n-1)}{n}\right]
\displaystyle =\lim_{n\to\infty}2\left[-2+2\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle =-4+8=4

\displaystyle \textbf{Question 22: }~\int_{0}^{2}(x^{2}+2)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=x^2+2,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 (x^2+2)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0+2)+(h^2+2)+(2h)^2+2+\cdots+((n-1)^2h^2+2)]
\displaystyle =\lim_{n\to\infty}h\,[2n+h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[2n+h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[2n+\frac{4}{n^2}\cdot\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}2\left[2+\frac{2}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle =4+\frac{8}{3}=\frac{20}{3}

\displaystyle \textbf{Question 23: }~\int_{0}^{4}(x+e^{2x})\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=4,\ f(x)=x+e^{2x},\ h=\frac{4-0}{n}=\frac{4}{n}
\displaystyle \text{Therefore, } I=\int_0^4 (x+e^{2x})\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0+e^0)+(h+e^{2h})+(2h+e^{4h})+\cdots+((n-1)h+e^{2(n-1)h})]
\displaystyle =\lim_{n\to\infty}h\,[h(1+2+3+\cdots+(n-1))+e^0+e^{2h}+e^{4h}+\cdots+e^{2(n-1)h}]
\displaystyle =\lim_{n\to\infty}h\left[h\frac{n(n-1)}{2}+\frac{(e^{2h})^{n}-1}{e^{2h}-1}\right]
\displaystyle =\lim_{n\to\infty}\frac{16}{n^2}\cdot\frac{n(n-1)}{2}+\lim_{n\to\infty}\frac{4}{n}\left[\frac{e^8-1}{e^{\frac{8}{n}}-1}\right]
\displaystyle =\lim_{n\to\infty}8\left(1-\frac{1}{n}\right)+(e^8-1)\lim_{n\to\infty}\frac{\frac{4}{n}}{e^{\frac{8}{n}}-1}
\displaystyle =8+\frac{e^8-1}{2}=\frac{15+e^8}{2}

\displaystyle \textbf{Question 24: }~\int_{0}^{2}(x^{2}+x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=x^2+x,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 (x^2+x)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0)+(h^2+h)+(2h)^2+2h+\cdots+((n-1)^2h^2+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[h^2(1^2+2^2+3^2+\cdots+(n-1)^2)+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[h^2\frac{n(n-1)(2n-1)}{6}+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[\frac{2(n-1)(2n-1)}{3n}+n-1\right]
\displaystyle =\lim_{n\to\infty}2\left[\frac{2}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+1-\frac{1}{n}\right]
\displaystyle =\frac{8}{3}+2=\frac{14}{3}

\displaystyle \textbf{Question 25: }~\int_{0}^{2}(x^{2}+2x+1)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=x^2+2x+1,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 (x^2+2x+1)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0^2+2\cdot0+1)+(h^2+2h+1)+(2h)^2+2(2h)+1+\cdots+((n-1)^2h^2+2(n-1)h+1)]
\displaystyle =\lim_{n\to\infty}h\,[n+h^2(1^2+2^2+3^2+\cdots+(n-1)^2)+2h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[n+h^2\frac{n(n-1)(2n-1)}{6}+2h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[n+\frac{2(n-1)(2n-1)}{3n}+2(n-1)\right]
\displaystyle =\lim_{n\to\infty}2\left[3+\frac{2}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)-\frac{2}{n}\right]
\displaystyle =6+\frac{8}{3}=\frac{26}{3}

\displaystyle \textbf{Question 26: }~\int_{0}^{3}(2x^{2}+3x+5)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=3,\ f(x)=2x^2+3x+5,\ h=\frac{3-0}{n}=\frac{3}{n}
\displaystyle \text{Therefore, } I=\int_0^3 (2x^2+3x+5)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0+0+5)+(2h^2+3h+5)+(2(2h)^2+3(2h)+5)+\cdots+(2(n-1)^2h^2+3(n-1)h+5)]
\displaystyle =\lim_{n\to\infty}h\,[5n+2h^2(1^2+2^2+3^2+\cdots+(n-1)^2)+3h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[5n+2h^2\frac{n(n-1)(2n-1)}{6}+3h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{3}{n}\left[5n+\frac{3(n-1)(2n-1)}{n}+\frac{9(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}3\left[5+3\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{9}{2}\left(1-\frac{1}{n}\right)\right]
\displaystyle =15+18+\frac{27}{2}=\frac{93}{3}

\displaystyle \textbf{Question 27: }~\int_{a}^{b}x\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=a,\ b=b,\ f(x)=x,\ h=\frac{b-a}{n}
\displaystyle \text{Therefore, } I=\int_a^b x\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(a)+f(a+h)+\cdots+f(a+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[a+(a+h)+(a+2h)+\cdots+(a+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[na+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[na+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{b-a}{n}\left[na+\frac{(b-a)(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\left[(b-a)a+\frac{(b-a)(b-a-\frac{b-a}{n})}{2}\right]
\displaystyle =(b-a)a+\frac{(b-a)^2}{2}
\displaystyle =\frac{2ab-2a^2+b^2+a^2-2ab}{2}
\displaystyle =\frac{b^2-a^2}{2}

\displaystyle \textbf{Question 28: }~\int_{0}^{5}(x+1)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=5,\ f(x)=x+1,\ h=\frac{5-0}{n}=\frac{5}{n}
\displaystyle \text{Therefore, } I=\int_0^5 (x+1)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0+1)+(h+1)+(2h+1)+\cdots+((n-1)h+1)]
\displaystyle =\lim_{n\to\infty}h\,[n+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[n+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{5}{n}\left[n+\frac{5(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}5\left[1+\frac{5}{2}\left(1-\frac{1}{n}\right)\right]
\displaystyle =5+\frac{25}{2}=\frac{35}{2}

\displaystyle \textbf{Question 29: }~\int_{2}^{3}x^{2}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=2,\ b=3,\ f(x)=x^2,\ h=\frac{3-2}{n}=\frac{1}{n}
\displaystyle \text{Therefore, } I=\int_2^3 x^2\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(2)+f(2+h)+f(2+2h)+\cdots+f(2+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[2^2+(2+h)^2+(2+2h)^2+\cdots+(2+(n-1)h)^2]
\displaystyle =\lim_{n\to\infty}h\,[4n+4h(1+2+3+\cdots+(n-1))+h^2(1^2+2^2+3^2+\cdots+(n-1)^2)]
\displaystyle =\lim_{n\to\infty}h\left[4n+4h\frac{n(n-1)}{2}+h^2\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{n\to\infty}\frac{1}{n}\left[4n+2(n-1)+\frac{(n-1)(2n-1)}{6n}\right]
\displaystyle =\lim_{n\to\infty}\left[6+\frac{1}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)-\frac{2}{n}\right]
\displaystyle =6+\frac{1}{3}=\frac{19}{3}

\displaystyle \textbf{Question 30: }~\int_{1}^{4}(x^{2}-x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=1,\ b=4,\ f(x)=x^2-x,\ h=\frac{4-1}{n}=\frac{3}{n}
\displaystyle \text{Therefore, } I=\int_1^4 (x^2-x)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(1^2-1)+((1+h)^2-(1+h))+((1+2h)^2-(1+2h))+\cdots+((1+(n-1)h)^2-(1+(n-1)h))]
\displaystyle =\lim_{n\to\infty}h\,[h^2(1^2+2^2+3^2+\cdots+(n-1)^2)+h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[h^2\frac{n(n-1)(2n-1)}{6}+h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{3}{n}\left[\frac{9(n-1)(2n-1)}{6n}+\frac{3(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}3\left[\frac{3}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{3}{2}\left(1-\frac{1}{n}\right)\right]
\displaystyle =9+\frac{9}{3}=\frac{38}{3}

\displaystyle \textbf{Question 31: }~\int_{0}^{2}(x^{2}-x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=0,\ b=2,\ f(x)=x^2-x,\ h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_0^2 (x^2-x)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(0)+f(h)+f(2h)+\cdots+f((n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(0^2-0)+(h^2-h)+(2h)^2-2h+\cdots+((n-1)^2h^2-(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[h^2(1^2+2^2+3^2+\cdots+(n-1)^2)-h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[h^2\frac{n(n-1)(2n-1)}{6}-h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[\frac{2(n-1)(2n-1)}{3n}-n+1\right]
\displaystyle =\lim_{n\to\infty}2\left[\frac{2}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)-1+\frac{1}{n}\right]
\displaystyle =\frac{8}{3}-2=\frac{2}{3}

\displaystyle \textbf{Question 32: }~\int_{1}^{3}(2x^{2}+5x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{where } h=\frac{b-a}{n}
\displaystyle \text{Here, } a=1,\ b=3,\ f(x)=2x^2+5x,\ h=\frac{3-1}{n}=\frac{2}{n}
\displaystyle \text{Therefore, } I=\int_1^3 (2x^2+5x)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,[(2\cdot1^2+5\cdot1)+(2(1+h)^2+5(1+h))+\cdots+(2(1+(n-1)h)^2+5(1+(n-1)h))]
\displaystyle =\lim_{n\to\infty}h\,[7n+2h^2(1^2+2^2+3^2+\cdots+(n-1)^2)+9h(1+2+3+\cdots+(n-1))]
\displaystyle =\lim_{n\to\infty}h\left[7n+2h^2\frac{n(n-1)(2n-1)}{6}+9h\frac{n(n-1)}{2}\right]
\displaystyle =\lim_{n\to\infty}\frac{2}{n}\left[7n+\frac{4(n-1)(2n-1)}{3n}+9n-9\right]
\displaystyle =\lim_{n\to\infty}2\left[16+\frac{4}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)-\frac{9}{n}\right]
\displaystyle =32+\frac{16}{3}=\frac{112}{3}

\displaystyle \textbf{Question 33: }~\int_{1}^{3}(3x^{2}+1)\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \int_a^b f(x)\,dx=\lim_{n\to\infty}h\,[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)]
\displaystyle \text{Here, } a=1,\ b=3,\ f(x)=3x^2+1
\displaystyle h=\frac{3-1}{n}=\frac{2}{n}\Rightarrow nh=2
\displaystyle \therefore \int_1^3 (3x^2+1)\,dx
\displaystyle =\lim_{n\to\infty}h\,[f(1)+f(1+h)+f(1+2h)+\cdots+f(1+(n-1)h)]
\displaystyle =\lim_{n\to\infty}h\,\{[3\cdot1^2+1]+[3(1+h)^2+1]+[3(1+2h)^2+1]+\cdots+[3(1+(n-1)h)^2+1]\}
\displaystyle =\lim_{n\to\infty}h\,\{3[1+(1+2h+h^2)+(1+4h+4h^2)+\cdots+(1+2(n-1)h+(n-1)^2h^2)]+n\}
\displaystyle =\lim_{n\to\infty}h\,\{3[n+2h(1+2+\cdots+(n-1))+h^2(1^2+2^2+\cdots+(n-1)^2)]+n\}
\displaystyle =\lim_{n\to\infty}h\left[4n+6\frac{n(n-1)}{2}h+3\frac{n(n-1)(2n-1)}{6}h^2\right]
\displaystyle =\lim_{n\to\infty}\left[4nh+3nh(nh-h)+\frac{(nh-h)nh(2nh-h)}{2}\right]
\displaystyle =4\cdot2+3\cdot2(2-h)+\frac{(2-h)\cdot2(4-h)}{2}
\displaystyle =8+12+8
\displaystyle =28

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.