The area bounded by the curve x=y2+4y with y- axis - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$\frac{28}{3}$
B
$\frac{32}{3}$
C
$\frac{13}{6}$
D
$\frac{23}{3}$

Solution:

The given curve is $x = y^{2} + 4 y$

This curve will pass through the origin
When $x = 0$ , $y^{2} + 4 y = 0 \Rightarrow y = 0, y = - 4$

The rough sketch of the given curve is as below

The required area is the area of the shaded region

The area of the shaded region is the area bounded by the curve $x = y^{2} + 4 y$ and lines $y = 0, y = - 4, y - a x i s$

Hence, the area of the shaded region is $A = |\int_{- 4}^{0} (y^{2} + 4 y) d y|$

It implies

$\begin{array}{rcl} A & = & |\int_{- 4}^{0} (y^{2} + 4 y) d y| \\ = & |{[\frac{y^{3}}{3} + 2 y^{2}]}_{- 4}^{0}| \\ = & |0 + \frac{64}{3} - 32| \\ = & \frac{32}{3} \end{array}$

Therefore, the area of the required region is $\frac{32}{3}$

The area bounded by the curve $x = y^{2} + 4 y$ with $y -$ axis

Solution:

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