If -01 e -x2 dx - a and -01 x 2 e -x2 dx - A - e - aB- then A - B is-

I = nbsp; ∫_0^1x^2 nbsp; e^ x^2dx ⇒ I= 12 nbsp; nbsp; nbsp;∫_0^1x( 2x nbsp; e^ x^2)dx Let u=x and v= 2x nbsp; e^ x^2 nbsp; Now, we use integra ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

If ∫01 e -x2 ...

Question

If $1 \int 0 e^{- x^{2}} d x = a$ and $1 \int 0 x^{2} e^{- x^{2}} d x = \frac{A}{e} + a B$ , then $A + B$ is .

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Solution

$I = 1 \int 0 x^{2} e^{- x^{2}} d x \Rightarrow I = \frac{- 1}{2} 1 \int 0 x (- 2 x e^{- x^{2}}) d x$
Let $u = x$ and $v = - 2 x e^{- x^{2}}$
Now, we use integration by parts.
Putting $e^{- x^{2}} = t \Rightarrow d t = - 2 x e^{- x^{2}} d x$
$\Rightarrow I = \frac{- 1}{2} ⎛ ⎜ ⎝ {[x e^{- x^{2}}]}_{0}^{- x^{2}} - 1 \int 0 e^{- x^{2}} d x ⎞ ⎟ ⎠ \Rightarrow I = \frac{- 1}{2} (e^{- 1} - a) \Rightarrow I = \frac{- 1}{2 e} + \frac{a}{2} = \frac{A}{e} + a B$

Hence, $A + B = 0$

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Similar questions

Q. Let

I_{1} = \int_{0}^{1} e^{- x^{2}} d x, I_{2} = \int_{0}^{1} e^{- x} {cos}^{2} x d x

and

I_{3} = \int_{0}^{1} e^{- x^{2}} {cos}^{2} x d x .

Then

If I is the greatest of the definite integrals
$I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x, I_{2} = \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x$
$I_{3} = \int_{0}^{1} e^{- x^{2}} d x, I_{4} = \int_{0}^{1} e^{\frac{- x^{2}}{2}} d x, t h e n$

Q. Consider the integrals

I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x

I_{2} = \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x, I_{3} = \int_{0}^{1} e^{- x^{2}} d x

and

I_{4} = \int_{0}^{1} e^{- (1 / 2) x^{2}} d x

. The greatest of these integrals

I

Q. Assertion :Let

I_{1} = \int_{0}^{1} e^{- x} {cos}^{2} x d x

I_{2} = \int_{0}^{1} e^{- x^{2}} {cos}^{2} x d x

I_{3} = \int_{0}^{1} e^{- x^{2}} d x

I_{4} = \int_{0}^{1} e^{\frac{- x^{2}}{2}} d x

then greatest of these integrals is

I_{4}

Reason:

\int_{a}^{b} f (x) d x < \int_{a}^{b} g (x) \forall f (x) \geq g (x)

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If 1∫0e−x2 dx=a and 1∫0x2e−x2 dx=Ae+aB, then A+B is .

I=1∫0x2e−x2dx⇒I=−121∫0x(−2xe−x2)dx Let u=x and v=−2xe−x2 Now, we use integration by parts. Putting e−x2=t⇒dt=−2xe−x2dx ⇒I=−12⎛⎜⎝[xe−x2]10−1∫0e−x2 dx⎞⎟⎠⇒I=−12(e−1−a)⇒I=−12e+a2=Ae+aB Hence, A+B=0

If $1 \int 0 e^{- x^{2}} d x = a$ and $1 \int 0 x^{2} e^{- x^{2}} d x = \frac{A}{e} + a B$ , then $A + B$ is .