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Let $f(x)$ be a function satisfying $f'(x)=f(x)$ with $f(0)=1\; and \; g(x)$ be a function that satisfies $f(x)+ g(x)=x^{2}.$ Then the value of the integral $\int_{0}^{1}f(x)g(x)dx\; is$

Option 1)
$e+\frac{e^{2}}{2}-\frac{3}{2}\;$

Option 2)
$\; e-\frac{e^{2}}{2}-\frac{3}{2}\;$

Option 3)
$\; e+\frac{e^{2}}{2}+\frac{5}{2}\;$

Option 4)
$\; e-\frac{e^{2}}{2}-\frac{5}{2}$

Answers (1)

best_answer

As learnt in

Integration By PARTS -

Let $u$ and $v$ are two functions then

$\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx$

- wherein

Where $u$ is the Ist function $v$ is he IInd function

$f(x)+g(x)=x^{2}$

$f'(x)=f(x)$ is possible when f(x)=e^x

Thus, $g(x)=(x^{2}-e^{x})$

$I=\int_{0}^{1} e^{x} (x^{2}-e^{x})dx$

$=\int_{0}^{1}x^{2}e^{x}dx-\int_{0}^{1}e^{2x}dx$

$=\left [ x^{2}e^{x}-2xe^{x}+2e^{x}-\frac{e^{2x}}{2}\right ]_{0}^{1}$

$=e-2e+2e-\frac{e^{2}}{2}-(0-0+2-\frac{1}{2})$

$=e-\frac{e^{2}}{2}-\frac{3}{2}$

Option 1)

$e+\frac{e^{2}}{2}-\frac{3}{2}\;$

This option is incorrect

Option 2)

$\; e-\frac{e^{2}}{2}-\frac{3}{2}\;$

This option is correct

Option 3)

$\; e+\frac{e^{2}}{2}+\frac{5}{2}\;$

This option is incorrect

Option 4)

$\; e-\frac{e^{2}}{2}-\frac{5}{2}$

This option is incorrect

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