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

For the following diagram. Find the area of the shaded region.

Accepted Answer

The diagram shows part of the graph of $y = - (x + 1) (x - 3)$

Now, the curve intersects the $x -$ axis i.e., $y = 0$

Therefore, $- (x + 1) (x - 3) = 0$

$\Rightarrow x + 1 = 0 and (x - 3) = 0$

$\Rightarrow x = - 1 and x = 3$

Use: The area included between $y = f (x), x - axis$ and ordinates $x = a; x = b$ is given by $\int_{a}^{b} y d x = \int_{a}^{b} f (x) d x$ .

Now, the required area of the shaded region $A = \int_{- 1}^{3} [- (x + 1) (x - 3)] d x$

$\Rightarrow A = \int_{- 1}^{3} [- (x^{2} - 3 x + x - 3)] d x$

$\Rightarrow A = \int_{- 1}^{3} [- x^{2} + 2 x + 3)] d x$

$\Rightarrow A = {(- \frac{x^{3}}{3} + 2 \times \frac{x^{2}}{2} + 3 x)}_{- 1}^{3}$

$\Rightarrow A = {(- \frac{x^{3}}{3} + x^{2} + 3 x)}_{- 1}^{3}$

$\Rightarrow A = (- \frac{{(3)}^{3}}{3} + {(3)}^{2} + 3 (3)) - (- \frac{{(- 1)}^{3}}{3} + {(- 1)}^{2} + 3 (- 1))$

$\Rightarrow A = (- \frac{27}{3} + 9 + 9) - (\frac{1}{3} + 1 - 3)$

$\Rightarrow A = (9) - (\frac{1}{3} - 2)$

$\Rightarrow A = (9) - (\frac{- 5}{3})$

$\Rightarrow A = \frac{32}{3}$ sq.units

Hence, the area of the required shaded region is $\frac{32}{3}$ sq.units.

For the following diagram. Find the area of the shaded region.

Important Questions on Application of Integrals

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