The area enclosed between the graph of y = x3 and the lines x = 0, y = 1, y = 8 is

Question

The area enclosed between the graph of y = x3 and the lines x = 0, y = 1, y = 8 is (A) (45/4) (B) 14 (C) 7 (D) none of these

Tardigrade Admin · Accepted Answer

Given curve is y = x3 or x = y1 /3 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/15b1cebca944f6556c695159f35dd4c3-.png" /> ∴ Required area=∫ limits18 y1/ 3 dy=[(y4 /3/4 /3)]18 =(3/4)[84 /3-14/ 3] =(3/4)×(16-1) =(3/4)×15 =(45/4) sq. units

Q. The area enclosed between the graph of $y = x^{3}$ and the lines $x = 0$ , $y = 1$ , $y = 8$ is

$\frac{45}{4}$

$14$

$7$

none of these

Given curve is $y = x^{3}$
or $x = y^{1/3}$

$∴$ Required area $= 1 \int 8 y^{1/3} d y = [\frac{y ^{4/3}}{4/3}]_{1}^{8}$
$= \frac{3}{4} [8^{4/3} - 1^{4/3}]$
$= \frac{3}{4} \times (16 - 1)$
$= \frac{3}{4} \times 15$
$= \frac{45}{4}$ sq. units

Q. The area enclosed between the graph of y=x3 and the lines x=0, y=1, y=8 is

445​

14

7

none of these

Given curve is y=x3 or x=y1/3 ∴ Required area=1∫8​y1/3dy=[4/3y4/3​]18​ =43​[84/3−14/3] =43​×(16−1) =43​×15 =445​ sq. units

Q. The area enclosed between the graph of $y = x^{3}$ and the lines $x = 0$ , $y = 1$ , $y = 8$ is

$\frac{45}{4}$

$14$

$7$

Given curve is $y = x^{3}$
or $x = y^{1/3}$

$∴$ Required area $= 1 \int 8 y^{1/3} d y = [\frac{y ^{4/3}}{4/3}]_{1}^{8}$
$= \frac{3}{4} [8^{4/3} - 1^{4/3}]$
$= \frac{3}{4} \times (16 - 1)$
$= \frac{3}{4} \times 15$
$= \frac{45}{4}$ sq. units