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Question

Mathematics Question on integral

Evaluate the definite integral as limit of sums: ∫−11exdx∫^1_{-1} e^xdx

Answer

The correct answer is:=(e−1e)=(e-\frac{1}{e})
Let I=∫−11exdx...(1)I=∫^1_{-1} e^xdx...(1)
It is known that,
∫abƒ(x)dx=(b−a)limn→∞1n[ƒ(a)+ƒ(a+h)...ƒ((a+(n−1)h]∫^b_a ƒ(x)dx=(b-a)\underset{n→∞}{lim}\frac{1}{n}[ƒ(a)+ƒ(a+h)...ƒ((a+(n-1)h],where h=b−anh=\frac{b-a}{n}
Here,a=−1,b=1,a=-1,b=1,and ƒ(x)=exƒ(x)=e^x
∴h=1+1n=2n∴h=\frac{1+1}{n}=\frac{2}{n}
∴I=(1+1)limn→∞1n[ƒ(−1)+ƒ(−1+2n)+ƒ(−1+2.2n)+...+ƒ(−1+(n−1)2n)]∴I=(1+1)\underset{n→∞}{lim}\frac{1}{n}[ƒ(-1)+ƒ(-1+\frac{2}{n})+ƒ(-1+2.\frac{2}{n})+...+ƒ(-1+\frac{(n-1)2}{n})]
=2limn→∞1n[e−1+e(−1+2n)+e(−1+22n)+...e(−1+(n−1)2n)]=2\underset{n→∞}{lim}\frac{1}{n}[e^{-1}+e^{(-1+\frac{2}{n})}+e^{(-1+2\frac{2}{n})}+...e^{(-1+(n-1)\frac{2}{n})}]
=2limn→∞1n[e−11+e2n+e4n+e6n+e(n−1)2n]=2\underset{n→∞}{lim}\frac{1}{n}[e^{-1}{1+e^{\frac{2}{n}}+e^{\frac{4}{n}}+e^{\frac{6}{n}}+e^{(n-1)\frac{2}{n}}}]
=2limn→∞e−1n[e2nn−1e2n−1]=2\underset{n→∞}{lim}\frac{e^{-1}}{n}[\frac{e^{\frac{2n}{n}-1}}{e^{\frac{2}{n}-1}}]
=e−1×2limn→∞1n[e2−1e2n−1]=e^{-1}\times2\underset{n→∞}{lim}\frac{1}{n}[\frac{e^{2-1}}{e^{\frac{2}{n}-1}}]
=e−1×2(e2−1)lim2n→0(e2n−12n)×2=\frac{e^{-1}×2(e^2-1)}{\underset{\frac{2}{n}→0}{lim}(\frac{e^{\frac{2}{n}-1}}{\frac{2}{n}})\times2}
=e−1[2(e2−1)2]      [limh→0(eh−1h)=1]=e^{-1}[\frac{2(e^2-1)}{2}]\,\,\,\,\,\, [\underset{h→0}{lim}(\frac{e^h-1}{h})=1]
=e2−1e=\frac{e^2-1}{e}
=(e−1e)=(e-\frac{1}{e})