Question

Evaluate the definite integral as limit of sums: \(∫^1_{-1} e^xdx\)

Answer 1

The correct answer is:\(=(e-\frac{1}{e})\)
Let \(I=∫^1_{-1} e^xdx...(1)\)
It is known that,
\(∫^b_a ƒ(x)dx=(b-a)\underset{n→∞}{lim}\frac{1}{n}[ƒ(a)+ƒ(a+h)...ƒ((a+(n-1)h]\),where \(h=\frac{b-a}{n}\)
Here,\(a=-1,b=1,\)and \(ƒ(x)=e^x\)
\(∴h=\frac{1+1}{n}=\frac{2}{n}\)
\(∴I=(1+1)\underset{n→∞}{lim}\frac{1}{n}[ƒ(-1)+ƒ(-1+\frac{2}{n})+ƒ(-1+2.\frac{2}{n})+...+ƒ(-1+\frac{(n-1)2}{n})]\)
\(=2\underset{n→∞}{lim}\frac{1}{n}[e^{-1}+e^{(-1+\frac{2}{n})}+e^{(-1+2\frac{2}{n})}+...e^{(-1+(n-1)\frac{2}{n})}]\)
\(=2\underset{n→∞}{lim}\frac{1}{n}[e^{-1}{1+e^{\frac{2}{n}}+e^{\frac{4}{n}}+e^{\frac{6}{n}}+e^{(n-1)\frac{2}{n}}}]\)
\(=2\underset{n→∞}{lim}\frac{e^{-1}}{n}[\frac{e^{\frac{2n}{n}-1}}{e^{\frac{2}{n}-1}}]\)
\(=e^{-1}\times2\underset{n→∞}{lim}\frac{1}{n}[\frac{e^{2-1}}{e^{\frac{2}{n}-1}}]\)
\(=\frac{e^{-1}×2(e^2-1)}{\underset{\frac{2}{n}→0}{lim}(\frac{e^{\frac{2}{n}-1}}{\frac{2}{n}})\times2}\)
\(=e^{-1}[\frac{2(e^2-1)}{2}]\,\,\,\,\,\, [\underset{h→0}{lim}(\frac{e^h-1}{h})=1]\)
\(=\frac{e^2-1}{e}\)
\(=(e-\frac{1}{e})\)