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)
P Plabita

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)=ex

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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