The area bounded by $y = 3 x$ and $y = x^{2}$ is (in square units)

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $4.5$
$y = 3 x$ and $y = x^{2} i n s q u n i t s$
$3 x = x^{2}$
$x = 3 a n d x = 0$ are the points of intersection
$3 \int 0 (3 x - x^{2}) d x$
$= \frac{3 x^{2}}{2} |_{0}^{3} - \frac{x^{3}}{3} |_{0}^{3}$

$= 3 \frac{9}{2} - \frac{27}{3}$

$= \frac{27}{2} - 9 = \frac{9}{2} = 4.5 s q u n i t s$

Suggest Corrections

Similar questions

y = x^{3}, y = x^{2}

x = 1, x = 2

x = - 1, x = 2, y = x^{2} + 1

y = 2 x - 2

x^{2} = 4 y, y = 1, y = 4

y + 2 x^{2} = 0

y + 3 x^{2} = 1

f (x) = x^{2} - 3 x + 2 \forall x ϵ R

f (x)

Join BYJU'S Learning Program