Let f(x) be a function satisfying f’(x) = f(x) with f(0) = 1 and g be the function satisfying f(x) + g(x) = x2. The value of the integral 0∫1 f(x)g(x)dx isoptions e – ½ e2 – 5/2 e – e2 – 3 ½ (e-3) e – ½ e2- 3/2

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Integral Calculus> Let f(x) be a function satisfying f’(x) =...

Let f(x) be a function satisfying f’(x) = f(x) with f(0) = 1 and g be the function satisfying f(x) + g(x) = x². The value of the integral ₀∫¹ f(x)g(x)dx is
options

e – ½ e² – 5/2

e – e² – 3

½ (e-3)

e – ½ e²- 3/2

Aditya Kartikeya , 11 Years ago

Grade 10

1 Answers

Jitender Singh

Ans:
Hello Student,
Please find answer to your question below

$f'(x) = f(x)$
$\frac{df(x)}{dx} = f(x)$
$\int \frac{df(x)}{f(x)} = \int dx$
$ln[f(x)] = x + c$
$f(x) = e^{x+c}$
$f(0) = e^{c}$
$1 = e^{c}$
$f(x) = e^{x}$
$f(x) + g(x) = x^2$
$e^{x} + g(x) = x^2$
$g(x) = x^2 - e^{x}$
$I = \int_{0}^{1}f(x)g(x)dx$
$I = \int_{0}^{1}e^{x}(x^2-e^{x})dx$
$I = \int_{0}^{1}x^2e^{x}dx-\int_{0}^{1}e^{2x}dx$
$I = (x^2e^{x}-2xe^{x}+2e^{x})_{0}^{1} - (\frac{e^{2x}}{2})_{0}^{1}$
$I = (e-2e+2e)_{0}^{1} - (\frac{e^{2}-1}{2})$
$I = (e-2e+2e-2) - (\frac{e^{2}-1}{2})$
$I = e-2 - \frac{e^{2}}{2}+\frac{1}{2}$
$I = e- \frac{e^{2}}{2}-\frac{3}{2}$
Option (4)

Last Activity: 11 Years ago