MHT CET · Maths · Indefinite Integration
If \(\int \mathrm{e}^{x^2}. x^3 \mathrm{~d} x=\mathrm{e}^{x^2} \mathrm{f}(x)+\mathrm{c}\) and \(\mathrm{f}(\mathrm{I})=0\) (where c is a constant of integration), then the value of \(\mathrm{f}(x)\) is
- A \(\frac{x-1}{2}\)
- B \(\frac{x^2+1}{2}\)
- C \(\frac{x+1}{2}\)
- D \(\frac{x^2-1}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{x^2-1}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text {Let } \mathrm{I}=\int \mathrm{e}^{\mathrm{x}^2} \cdot x^3 \mathrm{~d} x \\
& \text {Put } x^2=\mathrm{t} \\
& \Rightarrow 2 x \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad I=\frac{1}{2} \int \mathrm{e}^{\mathrm{t}} \cdot \mathrm{t} \mathrm{dt} \\
& =\frac{1}{2}\left(t \cdot e^t-\int 1 \cdot e^t\right) \\
& =\frac{1}{2}\left(\mathrm{te}^{\mathrm{t}}-\mathrm{e}^{\mathrm{t}}\right)+\mathrm{c} \\
& =\frac{1}{2} \mathrm{e}^{\mathrm{t}}(\mathrm{t}-1)+\mathrm{c}=\frac{1}{2} \mathrm{e}^{x^2}\left(x^2-1\right)+\mathrm{c} \\
& \therefore \quad \mathrm{f}(x)=\frac{x^2-1}{2}
\end{aligned}\)
& \text {Let } \mathrm{I}=\int \mathrm{e}^{\mathrm{x}^2} \cdot x^3 \mathrm{~d} x \\
& \text {Put } x^2=\mathrm{t} \\
& \Rightarrow 2 x \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad I=\frac{1}{2} \int \mathrm{e}^{\mathrm{t}} \cdot \mathrm{t} \mathrm{dt} \\
& =\frac{1}{2}\left(t \cdot e^t-\int 1 \cdot e^t\right) \\
& =\frac{1}{2}\left(\mathrm{te}^{\mathrm{t}}-\mathrm{e}^{\mathrm{t}}\right)+\mathrm{c} \\
& =\frac{1}{2} \mathrm{e}^{\mathrm{t}}(\mathrm{t}-1)+\mathrm{c}=\frac{1}{2} \mathrm{e}^{x^2}\left(x^2-1\right)+\mathrm{c} \\
& \therefore \quad \mathrm{f}(x)=\frac{x^2-1}{2}
\end{aligned}\)
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