The area enclosed between the curves y = x3 and y = √x is, (in sq unit) :

Question

The area enclosed between the curves y = x3 and y = √x is, (in sq unit) : (A) (5/3) (B) (5/4) (C) (5/12) (D) (12/5)

Tardigrade Admin · Accepted Answer

Given curves are y = x3 and y = √x From the above equation one can found that: x6 = x ⇒ x = 0, 1 Required area = ∫10 ( √x - x3 )dx = [(x3/2/3). 2 - (x4/4)]10 = (2/3) - (1/4) = (5/12) Sq. unit.

Q. The area enclosed between the curves $y = x^{3}$ and $y = x$ is, (in sq unit) :

$\frac{5}{3}$

$\frac{5}{4}$

$\frac{5}{12}$

$\frac{12}{5}$

Given curves are $y = x^{3}$ and $y = x$
From the above equation one can found that:
$x^{6} = x \Rightarrow x = 0, 1$
Required area $= \int_{0}^{1} (x - x^{3}) d x$
$= [\frac{x ^{3/2}}{3} .2 - \frac{x ^{4}}{4}]_{0}^{1} = \frac{2}{3} - \frac{1}{4}$
$= \frac{5}{12}$ Sq. unit.

Q. The area enclosed between the curves y=x3 and y=x​ is, (in sq unit) :

35​

45​

125​

512​