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Question

Mathematics Question on integral

Evaluate ∫01e2−3xdx∫^1_0e^{2-3x}dx as a limit of a sum.

Answer

The correct answer is:=13(e2−1e)=\frac{1}{3}(e^2-\frac{1}{e})
Let I=∫01e2−3xdxI=∫^1_0e^{2-3x}dx
It is known that,
∫abƒ(x)dx=(b−a)limn→∞1n[ƒ(a)+ƒ(a+h)+...+ƒ(a+(n−1)h)]∫^b_aƒ(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=0,b=1a=0,b=1,and ƒ(x)=e2−3xƒ(x)=e^{2-3x}
⇒h=1−0n=1n⇒h=\frac{1-0}{n}=\frac{1}{n}
∴∫01e2−3xdx=(1−0)limn→∞[ƒ(0)+ƒ(0+h)+...+ƒ(0+(n−1)h)]∴∫^1_0e^{2-3x}dx=(1-0)\underset{n→∞}{lim}[ƒ(0)+ƒ(0+h)+...+ƒ(0+(n-1)h)]
=limn→∞[e2+e2−3h+...e2−3(n−1)h]=\underset{n→∞}{lim}[e^2+e^{2-3h}+...e^{2-3(n-1)h}]
=limn→∞1n[e2[1+e−3h+e−6h+e−9h+...e−3(n−1)h]]=\underset{n→∞}{lim}\frac{1}{n}[e^2[1+e^{-3h}+e^{-6h}+e^{-9h}+...e^{-3(n-1)h}]]
=limn→∞1n[e2[1−(e−3h)n1−(e−3h)]]=\underset{n→∞}{lim}\frac{1}{n}[\frac{e^2[1-(e^{-3h})^n}{1-(e^{-3h})}]]
=limn→∞1n[e2(1−e−3)(1−e−3n)]=\underset{n→∞}{lim}\frac{1}{n}[e^2\frac{(1-e^{-3})}{(1-e^{\frac{-3}{n}})}]
=e2(e−3−1)limn→∞1n[1e−3n−1]=e^2(e^{-3}-1)\underset{n→∞}{lim}\frac{1}{n}\bigg[\frac{1}{e^{\frac{-3}{n}}-1}\bigg]
=e2(e−3−1)limn→∞(−13)[−3ne−3n−1]=e^2(e^{-3}-1)\underset{n→∞}{lim}(\frac{-1}{3})[\frac{\frac{-3}{n}}{e^{\frac{-3}{n}}-1}]
=e2(e−3−1)3limn→∞[−3ne−3n−1]=\frac{e^2(e^{-3}-1)}{3}\underset{n→∞}{lim}[\frac{\frac{-3}{n}}{e^{\frac{-3}{n}}-1}]
=−e2(e−3−1)3(1)     [limn→∞xex−1]=\frac{-e^2(e^{-3}-1)}{3}(1)\,\,\,\,\, [\underset{n→∞}{lim}\frac{x}{e^x-1}]
=−e−1+e23=\frac{-e^{-1}+e^2}{3}
=13(e2−1e)=\frac{1}{3}(e^2-\frac{1}{e})