Name: Simple Curve Sketching
Uploaded: 2019-07-19
Duration: 3 min 58 s
Description: This video gives us an introduction to simple curve sketching. Take out your notebook and pen, and start sketching simple curves while watching this video to...

Question

Draw a rough sketch and find the area bounded by the curve

x^{2} = y

and

x + y = 2

.

Accepted Answer

The given curves are $x^{2} = y$ and $x + y = 2$ .

$x^{2} = y$ is an upward parabola with vertex at origin.

Now finding the intersection points of given curves,

$x + y = 2$

$\Rightarrow y = 2 - x$

$\Rightarrow x^{2} = 2 - x$

$\Rightarrow x^{2} + x - 2 = 0$

$\Rightarrow x^{2} + 2 x - x - 2 = 0$

$\Rightarrow x (x + 2) - 1 (x + 2) = 0$

$\Rightarrow (x - 1) (x + 2) = 0$

$\Rightarrow x = 1$ and $x = - 2$

Now, at $x = 1$ , $y = 2 - 1 = 1$

At $x = - 2$ , $y = 2 + 2 = 4$

Thus, the points of intersection are $(1, 1)$ and $(- 2, 4)$ .

The required area of shaded region $= \int_{- 2}^{1} (2 - x) d x - \int_{- 2}^{1} x^{2} d x$

$= {[2 x - \frac{x^{2}}{2}]}_{- 2}^{1} - {[\frac{x^{3}}{3}]}_{- 2}^{1}$

$= [2 \times 1 - \frac{1^{2}}{2} - (2 \times (- 2) - \frac{{(- 2)}^{2}}{2})] - [\frac{1^{3}}{3} - \frac{{(- 2)}^{3}}{3}]$

$= [2 - \frac{1}{2} + 4 + 2] - [\frac{1}{3} + \frac{8}{3}]$

$= \frac{15}{2} - 3$

Required Area $= \frac{9}{2}$ square units

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

Important Questions on Applications of Integrals

Learn and Prepare for any exam you want !

Draw a rough sketch and find the area bounded by the curve x2=y and x+y=2.

Important Questions on Applications of Integrals

Learn and Prepare for any exam you want !

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