Definite IntegrationmediumFree

With I₁=∫x²/((2−x³)e^{x³})dx and I₂=∫eˣ/(1+x)dx over [0,1], put x³=z then z=1−y to get I₁=(1/3e)I₂, so I₁:I₂ = 1:3e.

JEE Maths question with a full step-by-step solution.

Question

I_{1}=\displaystyle\int_{0}^{1}\dfrac{x^{2}}{\big(2-x^{3}\big)\,e^{x^{3}}}\,dx

and

I_{2}=\displaystyle\int_{0}^{1}\dfrac{e^{x}}{1+x}\,dx

, then

I_{1}:I_{2}

equals

1:3

3:1

1:3e

correct

3e:1

Solution

Step 1: Substitute

x^{3}=z

I_{1}

, so

3x^{2}\,dx=dz

and

x^{2}\,dx=\dfrac{dz}{3}

, with

z:0\to 1

I_{1}=\dfrac13\int_{0}^{1}\dfrac{dz}{(2-z)\,e^{z}}=\dfrac13\int_{0}^{1}\dfrac{e^{-z}}{2-z}\,dz.

Step 2: Substitute

z=1-y

, so

dz=-dy

and

2-z=1+y

; the limits

z:0\to1

become

y:1\to0

I_{1}=\dfrac13\int_{1}^{0}\dfrac{e^{-(1-y)}}{1+y}\,(-dy)=\dfrac13\int_{0}^{1}\dfrac{e^{\,y-1}}{1+y}\,dy.

Step 3: Factor out

e^{-1}

I_{1}=\dfrac{1}{3e}\int_{0}^{1}\dfrac{e^{\,y}}{1+y}\,dy=\dfrac{1}{3e}\,I_{2},

since

\displaystyle\int_{0}^{1}\dfrac{e^{y}}{1+y}\,dy=I_{2}

. Step 4: Form the ratio:

\dfrac{I_{1}}{I_{2}}=\dfrac{1}{3e}\ \Rightarrow\ I_{1}:I_{2}=1:3e.

Correct answer: (3)

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