Let Ι1= displaystyle ∫ 01 ex2 d x and Ι2= displaystyle ∫ 01 2 x2 ex2 d x , then the value of Ι1+Ι2 is equal to

Question

Let Ι1= displaystyle ∫ 01 ex2 d x and Ι2= displaystyle ∫ 01 2 x2 ex2 d x , then the value of Ι1+Ι2 is equal to (A) 1 (B) 2 (C) e (D) e2

Tardigrade Admin · Accepted Answer

We have, Ι1= displaystyle ∫ 01 ex2 d x and Ι2= displaystyle ∫ 01 2 x2 ex2 d x Now, Ι2= displaystyle ∫ 01 2 x2 ex2 d x = displaystyle ∫ 01(2 x ex2)xdx Using integration by parts, Ι2=[x ex2]01- displaystyle ∫ 01 ex2 d x Ι2+Ι1=e

Q. Let $I_{1} = \int_{0}^{1} e^{x^{2}} d x$ and $I_{2} = \int_{0}^{1} 2 x^{2} e^{x^{2}} d x$ , then the value of $I_{1} + I_{2}$ is equal to

$1$

$2$

$e$

$e^{2}$

We have, $I_{1} = \int_{0}^{1} e^{x^{2}} d x$ and $I_{2} = \int_{0}^{1} 2 x^{2} e^{x^{2}} d x$
Now, $I_{2} = \int_{0}^{1} 2 x^{2} e^{x^{2}} d x$
$= \int_{0}^{1} (2 x e^{x^{2}}) x d x$
Using integration by parts,
$I_{2} = [x e^{x^{2}}]_{0}^{1} - \int_{0}^{1} e^{x^{2}} d x$
$I_{2} + I_{1} = e$

Q. Let I1​=∫01​ex2dx and I2​=∫01​2x2ex2dx , then the value of I1​+I2​ is equal to

1

2

e

e2

We have, I1​=∫01​ex2dx and I2​=∫01​2x2ex2dx Now, I2​=∫01​2x2ex2dx =∫01​(2xex2)xdx Using integration by parts, I2​=[xex2]01​−∫01​ex2dx I2​+I1​=e

Q. Let $I_{1} = \int_{0}^{1} e^{x^{2}} d x$ and $I_{2} = \int_{0}^{1} 2 x^{2} e^{x^{2}} d x$ , then the value of $I_{1} + I_{2}$ is equal to

$1$

$2$

$e$

$e^{2}$

We have, $I_{1} = \int_{0}^{1} e^{x^{2}} d x$ and $I_{2} = \int_{0}^{1} 2 x^{2} e^{x^{2}} d x$
Now, $I_{2} = \int_{0}^{1} 2 x^{2} e^{x^{2}} d x$
$= \int_{0}^{1} (2 x e^{x^{2}}) x d x$
Using integration by parts,
$I_{2} = [x e^{x^{2}}]_{0}^{1} - \int_{0}^{1} e^{x^{2}} d x$
$I_{2} + I_{1} = e$