The area bounded by the curves y = x(x - 3)^2 | vedclass

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

Question

(D) To find the area bounded by the curves $y = x(x - 3)^2$ and $y = x$,we first find their points of intersection by setting $x(x - 3)^2 = x$.$x((x -

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) To find the area bounded by the curves $y = x(x - 3)^2$ and $y = x$,we first find their points of intersection by setting $x(x - 3)^2 = x$.$x((x - 3)^2 - 1) = 0$$x(x^2 - 6x + 9 - 1) = 0$$x(x^2 - 6x + 8) = 0$$x(x - 2)(x - 4) = 0$The points of intersection are $x = 0, 2, 4$.The area $A$ is given by the integral:$A = \int_{0}^{2} |x(x - 3)^2 - x| dx + \int_{2}^{4} |x - x(x - 3)^2| dx$For $x \in [0, 2]$,$x(x - 3)^2 \ge x$,so the integral is $\int_{0}^{2} (x^3 - 6x^2 + 8x) dx$.$= [\frac{x^4}{4} - 2x^3 + 4x^2]_{0}^{2} = (\frac{16}{4} - 2(8) + 4(4)) - 0 = 4 - 16 + 16 = 4$.For $x \in [2, 4]$,$x \ge x(x - 3)^2$,so the integral is $\int_{2}^{4} (x - (x^3 - 6x^2 + 9x)) dx = \int_{2}^{4} (-x^3 + 6x^2 - 8x) dx$.$= [-\frac{x^4}{4} + 2x^3 - 4x^2]_{2}^{4} = (-64 + 128 - 64) - (-4 + 16 - 16) = 0 - (-4) = 4$.Total Area $= 4 + 4 = 8 	ext{ sq. units}$.

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

Similar Questions

Which of the following is an example of an unsaturated fatty acid?

The value of electric potential at any point due to an electric dipole is

Which of the following is the most primitive ancestor of humans?

If $a = i + j + k$,$b = i + j$,$c = i$ and $(a \times b) \times c = \lambda a + \mu b$,then $\lambda + \mu = $

Which of the following compounds is not chiral?

Vedclass Test Series

Exam Paper Generator

Online Exam Module