The area bounded by the curve y=x2(x - 1)2 with the x -axis is k sq. units, then the value of 60k is equal to

Question

The area bounded by the curve y=x2(x - 1)2 with the x -axis is k sq. units, then the value of 60k is equal to (A) 1 (B) 2 (C) (1/2) (D) (1/4)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-hwcjdltjayr9.png" /> The required area is A= displaystyle ∫ 01 x2 (x - 1)2 d x or A= displaystyle ∫ 01 (x2 - x)2 d x= displaystyle ∫ 01 (x4 + x2 - 2 x3)dx =[(x5/5) + (x3/3) - (2 x4/4)]01 =(1/5)+(1/3)-(2/4)=(1/30)=k ∴ 60k=2

Q. The area bounded by the curve $y = x^{2} (x - 1)^{2}$ with the $x$ -axis is $k$ sq. units, then the value of $60 k$ is equal to

$1$

$2$

$\frac{1}{2}$

$\frac{1}{4}$

The required area is $A = \int_{0}^{1} x^{2} (x - 1)^{2} d x$
or $A = \int_{0}^{1} (x^{2} - x)^{2} d x = \int_{0}^{1} (x^{4} + x^{2} - 2 x^{3}) d x$
$= [\frac{x ^{5}}{5} + \frac{x ^{3}}{3} - \frac{2 x ^{4}}{4}]_{0}^{1}$
$= \frac{1}{5} + \frac{1}{3} - \frac{2}{4} = \frac{1}{30} = k$
$∴ 60 k = 2$

Q. The area bounded by the curve y=x2(x−1)2 with the x -axis is k sq. units, then the value of 60k is equal to

1

2

21​

41​

The required area is A=∫01​x2(x−1)2dx or A=∫01​(x2−x)2dx=∫01​(x4+x2−2x3)dx =[5x5​+3x3​−42x4​]01​ =51​+31​−42​=301​=k ∴60k=2

Q. The area bounded by the curve $y = x^{2} (x - 1)^{2}$ with the $x$ -axis is $k$ sq. units, then the value of $60 k$ is equal to

$1$

$2$

$\frac{1}{2}$

$\frac{1}{4}$

The required area is $A = \int_{0}^{1} x^{2} (x - 1)^{2} d x$
or $A = \int_{0}^{1} (x^{2} - x)^{2} d x = \int_{0}^{1} (x^{4} + x^{2} - 2 x^{3}) d x$
$= [\frac{x ^{5}}{5} + \frac{x ^{3}}{3} - \frac{2 x ^{4}}{4}]_{0}^{1}$
$= \frac{1}{5} + \frac{1}{3} - \frac{2}{4} = \frac{1}{30} = k$
$∴ 60 k = 2$