Find the area bounded by the curve y = x{e^{{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The required area $A$ is given by the definite integral: $A = \int_0^a y \, dx = \int_0^a x{e^{{x^2}}} \, dx$ Let $t = x^2$. Then,differentiating

Vedclass · Accepted Answer

$\frac{{{e^{{a^2}}} - 1}}{2} 	ext{ sq. unit}$

Vedclass · Answer

(B) The required area $A$ is given by the definite integral: $A = \int_0^a y \, dx = \int_0^a x{e^{{x^2}}} \, dx$ Let $t = x^2$. Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$. Changing the limits of integration: When $x = 0$,$t = 0^2 = 0$. When $x = a$,$t = a^2$. Substituting these into the integral: $A = \int_0^{a^2} {e^t} \cdot \frac{dt}{2} = \frac{1}{2} \int_0^{a^2} {e^t} \, dt$ Evaluating the integral: $A = \frac{1}{2} [e^t]_0^{a^2} = \frac{1}{2} (e^{a^2} - e^0) = \frac{e^{a^2} - 1}{2} 	ext{ sq. unit}$.

Find the area bounded by the curve $y = x{e^{{x^2}}}$,the $x$-axis,and the ordinates $x = 0$ and $x = a$.

Similar Questions

The area (in square units) of the region bounded by $x^2=8y$, $x=4$ and the $X$-axis is:

The area of the smaller segment cut off from the circle $x^2 + y^2 = 9$ by the line $x = 1$ is

Find the area of the region enclosed by the parabola $x^{2}=y,$ the line $y=x+2$ and the $x-$ axis.

The area bounded by the curve $y = x^2 + 2$, the $x$-axis, and the lines $x = 1$ and $x = 2$ is:

If $\int\limits_0^1 {(4x^3 - f(x))f(x)dx = \frac{4}{7}}$,then the area of the region bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = 1$ and $x = 2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module