Let x>0 be a fixed real number. Then the integral ∫_0^∞ e^-t|x-t| d t is equal to.

Correct answer: (A) x+2 e^-x-1. ∫_0^x e^-t|(x-t)| dt =∫_0^x e_11^-1(x-t) dt+∫_x^x e_11^-1(t-x) dt \-(x-t) e^-t-(-1) e^-t \_0^x- \-(x-t) e^-t+e^-t )_x =x+2 e^-x-1

Q54 / 90 · KVPY 2015 (SB/SX) · Mathematics

Let $\mathrm{x}>0$ be a fixed real number. Then the integral $\int_{0}^{\infty} e^{-t}|x-t| d t$ is equal to.

Correct answer
A
$x+2 e^{-x}-1$
B
$x-2 e^{-x}+1$
C
$x+2 e^{-x}+1$
D
$-x-2 e^{-x}+1$

Solution

$\int_{0}^{x} \mathrm{e}^{-t}|(\mathrm{x}-\mathrm{t})| \mathrm{dt}$ $=\int_{0}^{\mathrm{x}} \mathrm{e}_{11}^{-1}(\mathrm{x}-\mathrm{t}) \mathrm{dt}+\int_{\mathrm{x}}^{x} \mathrm{e}_{11}^{-1}(\mathrm{t}-\mathrm{x}) \mathrm{dt}$ $\left\{-(\mathrm{x}-\mathrm{t}) \mathrm{e}^{-t}-(-1) \mathrm{e}^{-\mathrm{t}}\right\}_{0}^{\mathrm{x}}-\left\{-(\mathrm{x}-\mathrm{t}) \mathrm{e}^{-\mathrm{t}}+\mathrm{e}^{-t}\right)_{\mathrm{x}}$ $=\mathrm{x}+2 \mathrm{e}^{-x}-1$

Hence, the correct answer is option: 'A'.

Let $\mathrm{x}>0$ be a fixed real number. Then the integral $\int_{0}^{\infty} e^{-t}|x-t| d t$ is equal to.

Solution

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