The area of the region bounded by the curves | vedclass

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

Question

(B) The area $A$ is given by the integral of the absolute difference between the curves: $A = \int_{0}^{2} |x^3 - x^2| \, dx$. First,find the intersec

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) The area $A$ is given by the integral of the absolute difference between the curves: $A = \int_{0}^{2} |x^3 - x^2| \, dx$. First,find the intersection points: $x^3 = x^2 \implies x^2(x-1) = 0$,so $x=0$ and $x=1$. In the interval $[0, 1]$,$x^2 \ge x^3$,so $|x^3 - x^2| = x^2 - x^3$. In the interval $[1, 2]$,$x^3 \ge x^2$,so $|x^3 - x^2| = x^3 - x^2$. Thus,$A = \int_{0}^{1} (x^2 - x^3) \, dx + \int_{1}^{2} (x^3 - x^2) \, dx$. Evaluating the first integral: $[\frac{x^3}{3} - \frac{x^4}{4}]_{0}^{1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$. Evaluating the second integral: $[\frac{x^4}{4} - \frac{x^3}{3}]_{1}^{2} = (\frac{16}{4} - \frac{8}{3}) - (\frac{1}{4} - \frac{1}{3}) = (4 - \frac{8}{3}) - (-\frac{1}{12}) = \frac{4}{3} + \frac{1}{12} = \frac{16+1}{12} = \frac{17}{12}$. Total area $A = \frac{1}{12} + \frac{17}{12} = \frac{18}{12} = \frac{3}{2}$.

The area of the region bounded by the curves $y=x^3$,$y=x^2$ and the lines $x=0$ and $x=2$ is

Similar Questions

Area of the region bounded by the curves $y = \sin x$ and $y = x$ between the lines $x = 0$ and $x = 2\pi$ is:

The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to: (in $/3$)

The area enclosed by the parabola ${y^2} = 4ax$ and the straight line $y = 2ax$ is

Find the area between the curves $y=x$ and $y=x^{2}$.

The polynomial $f(x)$ satisfies the condition $f(x + 1) = x^2 + 4x$. The area enclosed by $y = f(x - 1)$ and the curve $x^2 + y = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module