Name: Area Enclosed by a Curve and a Line: Illustration
Uploaded: 2019-07-19
Duration: 3 min 35 s
Description: How would you find the area enclosed between a parabola and a straight line? Watch the video to see the method as an application of definite integrals.

Question

Draw a rough sketch of the curve $y^{2} = 8 x$ and find the area of the region bounded by the curve $y^{2} = 8 x$ and the line $x = 2$ .

Accepted Answer

Let, $A B$ represent line $x = 2$

& $A O B$ represent the curve $y^{2} = 8 x$

We have to find area of $A O B$ .

Area of $A O B =$ $2 \int_{0}^{2} y d x$

So we need to calculate $\int_{0}^{2} y d x$

We know that $y^{2} = 8 x$

Taking square root on both sides,

$\Rightarrow$ $y = \pm 2 \sqrt{2} x$

Area of $A O B =$ $2 \int_{0}^{2} y d x = 2 \int_{0}^{2} 2 \sqrt{2 x} d x = 4 \sqrt{2} \int_{0}^{2} {(x)}^{\frac{1}{2}} d x$

As we know, $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$ .

Area of $A O B =$ $4 \sqrt{2} {[\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}]}_{0}^{2} = 4 \sqrt{2} {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{2} = \frac{8 \sqrt{2}}{3} {[x^{\frac{3}{2}}]}_{0}^{2} = \frac{8 \sqrt{2}}{3} {[{(2)}^{3}]}^{\frac{1}{2}} = \frac{8 \sqrt{2}}{3} [2 \sqrt{2}] = \frac{32}{3}$

Draw a rough sketch of the curve $y^{2} = 8 x$ and find the area of the region bounded by the curve $y^{2} = 8 x$ and the line $x = 2$ .

Important Questions on Applications of Integrals

Learn and Prepare for any exam you want !

Draw a rough sketch of the curve y2=8x and find the area of the region bounded by the curve y2=8x and the line x=2.

Important Questions on Applications of Integrals

Learn and Prepare for any exam you want !

Draw a rough sketch of the curve $y^{2} = 8 x$ and find the area of the region bounded by the curve $y^{2} = 8 x$ and the line $x = 2$ .