Let x -0 be a fixed real number- Then the integral -0- e - t - x - t - dt is equal to A- x-2 e-x-1B- x-2 e-x-1C- x-2 e-x-1D- -x-2 e-x-1

The correct option is B x+2e^ x 1∫_0^∞e^ t|x t|dt =∫_0^xe^ t(x t)dt+∫_x^∞e^ t(t x)dt Apply intregration by parts [ (x t)e^ t ( 1)e^ t]_0^x [ (x t)e^ t+e^ t]_x^ ...

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Byju's Answer

Standard XII

Mathematics

Area between Two Curves

Let x >0 be a...

Question

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

x + 2 e^{- x} - 1

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x - 2 e^{- x} + 1

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x + 2 e^{- x} + 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- x - 2 e^{- x} + 1

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Solution

The correct option is A $x + 2 e^{- x} - 1$
$\infty \int 0 e^{- t} | x - t | d t = x \int 0 e^{- t} (x - t) d t + \infty \int x e^{- t} (t - x) d t$
Apply intregration by parts
$[- (x - t) e^{- t} - (- 1) e^{- t}]_{0}^{x} - [- (x - t) e^{- t} + e^{- t}]_{x}^{\infty}$
$x + 2 e^{- x} - 1$

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Similar questions

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Let x>0 be a fixed real number. Then the integral ∞∫0e−t|x−t|dt is equal to

The correct option is A x+2e−x−1∞∫0e−t|x−t|dt=x∫0e−t(x−t)dt+∞∫xe−t(t−x)dt Apply intregration by parts [−(x−t)e−t−(−1)e−t]x0−[−(x−t)e−t+e−t]∞x x+2e−x−1

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

The correct option is A $x + 2 e^{- x} - 1$
$\infty \int 0 e^{- t} | x - t | d t = x \int 0 e^{- t} (x - t) d t + \infty \int x e^{- t} (t - x) d t$
Apply intregration by parts
$[- (x - t) e^{- t} - (- 1) e^{- t}]_{0}^{x} - [- (x - t) e^{- t} + e^{- t}]_{x}^{\infty}$
$x + 2 e^{- x} - 1$