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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
Let $x > 0$ be a fixed real number. Then, the integral $\int \limits_0^{\infty} e^{-t}|x-t| d t$ is equal to
  • ✓
    $x+2 e^{-x}-1$
  • B
    $x-2 e^{-x}+1$
  • C
    $x+2 e^{-x}+1$
  • D
    $-x-2 e^{-x}+1$
✓

Answer

Correct option: A.
$x+2 e^{-x}-1$
a
(a)

Let $I=\int \limits_0^{\infty} e^{-t}|x-t| d t, x > 0$

$\Rightarrow \quad I=\int \limits_0^x e^{-t}(x-t) d t+\int \limits_x^{\infty} e^{-t}(t-x) d t$

$\Rightarrow \quad I=\left[-(x-t) e^{-t}\right]_0^x+\int \limits_0^x \frac{d(x-t)}{d x} e^{-t} d t$

$+\left[\frac{(t-x) e^{-t}}{-1}\right]_x^{\infty}+\int \limits_x^{\infty} e^{-t} d t$

$\Rightarrow I=x+\left[e^{-t}\right]_0^x+\left[-e^{-t}\right]_x^{\infty}$

$\Rightarrow I=x+e^{-x}-1+e^{-x}$

$\Rightarrow I=x+2 e^{-x}-1$

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