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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
Let $f(x)$ be a function satisfying $f'(x) = f(x)$ with $f(0) = 1$ and $g(x)$ be the function satisfying $f(x) + g(x) = {x^2}.$ The value of integral $\int_0^1 {f(x)\,g(x)\,dx} $ is equal to
  • A
    $\frac{1}{4}(e - 7)$
  • B
    $\frac{1}{4}(e - 2)$
  • C
    $\frac{1}{2}(e - 3)$
  • ✓
    None of these
✓

Answer

Correct option: D.
None of these
d
(d) We have $f'(x) = f(x) \Rightarrow \frac{{f'(x)}}{{f(x)}} = 1$

==> $\log f(x) = x + \log c \Rightarrow f(x) = c{e^x}$

Since $f(0) = 1$,

therefore $1 = c{e^0} \Rightarrow c = 1$

Thus $f(x) = {e^x}$.

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

$\therefore \,\,\int_0^1 {f(x)g(x)dx = \int_0^1 {{e^x}({x^2}} - {e^x})dx} $

$ = e - \frac{1}{2}{e^2} - \frac{3}{2}$.

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