What is the area of the region $\left\{(x, y): 0 \leq y \leq x e^{x^2}, 0 \leq x \leq 1\right\}$ ?...

To find the area of the region $\left\{(x, y): 0 \leq y \leq x e^{x^2}, 0 \leq x \leq 1\right\}$, we need to integrate the function $x e^{x^2}$ from 0 to 1. We can achieve this using the method of substitution.Let $t = x^2$. Then,$ \frac{dt}{dx} = 2x \quad \Rightarrow \quad x = \frac{1}{2} \frac{dt}{dx} $Substituting this in, the integral becomes:$ \int_0^1 x e^{x^2} \, dx = \int_0^1 \frac{1}{2} \frac{dt}{dx} e^t \, dx $This simplifies to:$ = \frac{1}{2} \int_0^1 e^t \, dt $Evaluating the integral of $e^t$ from 0 to 1 gives:$ = \frac{1}{2} \left[ e^t \right]_0^1 = \frac{1}{2}(e - 1) $So, the area of the region is $\frac{1}{2}(e - 1)$.

Area Under The Curves - IAT-IISER Maths PYQs

To find the area of the region

%5Cleft%5C%7B(x%2C%20y)%3A%200%20%5Cleq%20y%20%5Cleq%20x%20e%5E%7Bx%5E2%7D%2C%200%20%5Cleq%20x%20%5Cleq%201%5Cright%5C%7D

, we need to integrate the function

x%20e%5E%7Bx%5E2%7D

from 0 to 1. We can achieve this using the method of substitution.Let

t%20%3D%20x%5E2

. Then,

%5Cfrac%7Bdt%7D%7Bdx%7D%20%3D%202x%20%5Cquad%20%5CRightarrow%20%5Cquad%20x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bdt%7D%7Bdx%7D

Substituting this in, the integral becomes:

%5Cint_0%5E1%20x%20e%5E%7Bx%5E2%7D%20%5C%2C%20dx%20%3D%20%5Cint_0%5E1%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bdt%7D%7Bdx%7D%20e%5Et%20%5C%2C%20dx

This simplifies to:

%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cint_0%5E1%20e%5Et%20%5C%2C%20dt

Evaluating the integral of

e%5Et

from 0 to 1 gives:

%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cleft%5B%20e%5Et%20%5Cright%5D_0%5E1%20%3D%20%5Cfrac%7B1%7D%7B2%7D(e%20-%201)

So, the area of the region is

%5Cfrac%7B1%7D%7B2%7D(e%20-%201)

Area Under The Curves

High-Yield Trend

Questions
1 MCQs

Official Solution

Area Under The Curves

High-Yield Trend

Questions 1 MCQs

Official Solution

Questions
1 MCQs