The area enclosed by the curves y = x^2,y = x | vedclass

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

Question

(D) The given curves are $y = x^2$ and $y = x^3$.The intersection point of $y = x^2$ and $y = x^3$ is found by setting $x^2 = x^3$,which gives $x^2(x

Vedclass · Accepted Answer

$4/3$

Vedclass · Answer

(D) The given curves are $y = x^2$ and $y = x^3$.The intersection point of $y = x^2$ and $y = x^3$ is found by setting $x^2 = x^3$,which gives $x^2(x - 1) = 0$,so $x = 0$ or $x = 1$.For $0  x^3$,and for $x > 1$,$x^3 > x^2$.Given $p > 1$,the area is the sum of the areas in the intervals $[0, 1]$ and $[1, p]$:Area $= \int_{0}^{1} (x^2 - x^3) \, dx + \int_{1}^{p} (x^3 - x^2) \, dx = \frac{1}{6}$.Calculating the first integral:$\int_{0}^{1} (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$.Calculating the second integral:$\int_{1}^{p} (x^3 - x^2) \, dx = \left[ \frac{x^4}{4} - \frac{x^3}{3} \right]_{1}^{p} = \left( \frac{p^4}{4} - \frac{p^3}{3} \right) - \left( \frac{1}{4} - \frac{1}{3} \right) = \frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{12}$.Summing the areas:$\frac{1}{12} + \frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{12} = \frac{1}{6}$.$\frac{p^4}{4} - \frac{p^3}{3} + \frac{2}{12} = \frac{1}{6}$.$\frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{6} = \frac{1}{6}$.$\frac{p^4}{4} - \frac{p^3}{3} = 0$.Multiplying by $12$:$3p^4 - 4p^3 = 0$.$p^3(3p - 4) = 0$.Since $p > 1$,we have $p = 4/3$.

The area enclosed by the curves $y = x^2$,$y = x^3$,$x = 0$,and $x = p$,where $p > 1$,is $1/6$. The value of $p$ is:

Similar Questions

The area bounded by the curves $y = x(x - 3)^2$ and $y = x$ is (in sq. units) :

The area bounded by the parabola $y^2=x$ and the line $x+y=2$ in the first quadrant is

The area bounded by the curves $y = |x| - 1$ and $y = -|x| + 1$ is

The area included between the parabolas $y^{2} = 5x$ and $x^{2} = 5y$ is

The area of the region $A = \{(x, y) : |\cos x - \sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2}\}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module