JEE MAIN - Mathematics (2003 - No. 19)
Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,} $$ is
$$e + {{{e^2}} \over 2} + {5 \over 2}$$
$$e - {{{e^2}} \over 2} - {5 \over 2}$$
$$e + {{{e^2}} \over 2} - {3 \over 2}$$
$$e - {{{e^2}} \over 2} - {3 \over 2}$$
Explanation
Given $$f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$$
Integrating log
$$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$
$$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$$
$$\therefore$$ $$\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx$$
$$ = \int\limits_0^1 {{e^x}} \left( {{x^2} - {e^x}} \right)dx$$
$$ = \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx$$
$$ = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1$$
$$ = e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]$$
$$ = e - {{{e^2}} \over 2} - {3 \over 2}$$
Integrating log
$$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$
$$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$$
$$\therefore$$ $$\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx$$
$$ = \int\limits_0^1 {{e^x}} \left( {{x^2} - {e^x}} \right)dx$$
$$ = \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx$$
$$ = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1$$
$$ = e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]$$
$$ = e - {{{e^2}} \over 2} - {3 \over 2}$$
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