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

Using integration, find the area of the triangle

A B C

whose vertices have coordinate

A (2, 5), B (4, 7)

and

C (6, 2)

Accepted Answer

First we find the equations of the sides of triangle $A B C$ by using two point form i.e., $y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$

The equation of $A B$ is $y - 5 = (\frac{7 - 5}{4 - 2}) (x - 2)$

$\Rightarrow y = x + 3$ …(i)

The equation of $B C$ is $y - 7 = (\frac{2 - 7}{6 - 4}) (x - 4)$

$\Rightarrow y = - \frac{1}{2} (5 x - 34)$ …(ii)

The equation of side $A C$ is $y - 5 = (\frac{2 - 5}{6 - 2}) (x - 2)$

$\Rightarrow y = - \frac{1}{4} (3 x - 26)$ …(iii)

Therefore, area of $△ A B C = \int_{2}^{4} y_{AB} d x + \int_{4}^{6} y_{BC} d x - \int_{2}^{6} y_{AC} d x$

$= \int_{2}^{4} (x + 3) d x + \int_{4}^{6} - \frac{1}{2} (5 x - 34) d x - \int_{2}^{6} - \frac{1}{4} (3 x - 26) d x$

$= {[\frac{x^{2}}{2} + 3 x]}_{2}^{4} - \frac{1}{2} {[\frac{5 x^{2}}{2} - 34 x]}_{4}^{6} + \frac{1}{4} {[\frac{3 x^{2}}{2} - 26 x]}_{2}^{6}$

$= [(\frac{16}{2} + 12) - (\frac{4}{2} + 6)] - \frac{1}{2} [(\frac{180}{2} - 204) - (\frac{80}{2} - 136)] + \frac{1}{4} [(\frac{108}{2} - 156) - (\frac{12}{2} - 52)]$

$= [(8 + 12) - (2 + 6)] - \frac{1}{2} [(90 - 204) - (40 - 136)] + \frac{1}{4} [(54 - 156) - (6 - 52)]$

$= 12 + \frac{1}{2} (18) - \frac{1}{4} (56) = 12 + 9 - 14 = 7 sq. units.$

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