MHT CET · Maths · Indefinite Integration
If \(\int e^{x^2} \cdot x^3 d x=e^{x^2} f(x)+C\)
(where \(C\) is a constant of integration)
and \(f(1)=0\), then value of \(f(2)\) will be
- A \(\frac{3}{2}\)
- B \(\frac{1}{2}\)
- C \(\frac{-3}{2}\)
- D \(\frac{-1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(\int e^{x^2 x^3}=\frac{1}{2} \int x^2 e^{x^2} 2 x d x=\frac{1}{2} \int t e^t d t \quad\left[\operatorname{let} x^2=t\right]\)
Integrating by parts
\(\begin{aligned} & =\frac{1}{2}\left[t \cdot e^t-e^t\right]+C \\ & =\frac{1}{2}\left[x^2 e^{x^2}-e^{x^2}\right]+C=e^{x^2} \cdot \frac{1}{2}\left(x^2-1\right)+C \\ & \Rightarrow f(x)=\frac{1}{2}\left(x^2-1\right) \Rightarrow f(2)=\frac{3}{2}\end{aligned}\)
Integrating by parts
\(\begin{aligned} & =\frac{1}{2}\left[t \cdot e^t-e^t\right]+C \\ & =\frac{1}{2}\left[x^2 e^{x^2}-e^{x^2}\right]+C=e^{x^2} \cdot \frac{1}{2}\left(x^2-1\right)+C \\ & \Rightarrow f(x)=\frac{1}{2}\left(x^2-1\right) \Rightarrow f(2)=\frac{3}{2}\end{aligned}\)
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