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Question

Mathematics Question on integral

Evaluate the definite integral as limit of sums: ∫04(x+e2x)dx∫^4_0(x+e^{2x})dx

Answer

The correct answer is:15+e82\frac{15+e^8}{2}
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-a/n
Here,a=0,b=4a=0,b=4,and ƒ(x)=x+e2xƒ(x)=x+e^{2x}
∴h=4−0n=4n∴h=\frac{4-0}{n}=\frac{4}{n}
⇒∫04(x+e2x)dx=(4−0)limn→∞1n[ƒ(0)+ƒ(h)+ƒ(2h)+...+ƒ((n−1)h)]⇒∫^4_0(x+e^{2x})dx=(4-0)\underset{n→∞}{lim}\frac{1}{n}[ƒ(0)+ƒ(h)+ƒ(2h)+...+ƒ((n-1)h)]
=4limn→∞1n[(0+e°)+(n+e2h)+(2h+e2.2h)+...+[(n−1)h+e2(n−1)h]]=4\underset{n→∞}{lim}\frac{1}{n}[(0+e°)+(n+e^{2h})+(2h+e^{2.2h})+...+[(n-1)h+e^{2(n-1)h}]]
=4limn→∞1n[1+(h+e2h)+(2h+e4h)+...+[(n−1)h+e2(n−1)h]]=4\underset{n→∞}{lim}\frac{1}{n}[1+(h+e^{2h})+(2h+e^{4h})+...+[(n-1)h+e^{2(n-1)h}]]
=4limn→∞1n[[h+2h+3h+...+(n−1)h]+(1+e2h+e4h+...+e2(n−1)h)]=4\underset{n→∞}{lim}\frac{1}{n}[[h+2h+3h+...+(n-1)h]+(1+e^{2h}+e^{4h}+...+e^{2(n-1)h})]
=4limn→∞1n[h[1+2+...(n−1)]+(e2hn−1e2h−1)]=4\underset{n→∞}{lim}\frac{1}{n}[h[1+2+...(n-1)]+\frac{(e^{2hn}-1}{e^{2h}-1)}]
=4limn→∞1n[(h(n−1)n)2+(e2hn−1e2h−1)]=4\underset{n→∞}{lim}\frac{1}{n}[\frac{(h(n-1)n)}{2}+(\frac{e^{2hn}-1}{e^{2h}-1})]
=4limn→∞1n[4n.(n−1)n2+(e8−1e8n−1)]=4\underset{n→∞}{lim}\frac{1}{n}[\frac{4}{n}.\frac{(n-1)n}{2}+(\frac{e^8-1}{e^{\frac{8}{n}}-1})]
=4(2)+4limn→∞(e8−1)(e8n−18n)8=4(2)+4\underset{n→∞}{lim}\frac{(e^8-1)}{(\frac{e^{\frac{8}{n}-1}}{\frac{8}{n}})8}
=8+e8−12=8+\frac{e^8-1}{2}
=15+e82=\frac{15+e^8}{2}