Integral calculus:Application

Leamir M. Sakal's picture
Submitted by Leamir M. Sakal on May 9, 2017 - 10:48pm

Problem: Find the area of the region bounded by the graphs as shown in the picture, given that the 4a=2.problem1.png

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Area by integration Integral application
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Jhun Vert's picture

Submitted by Jhun Vert on June 7, 2017 - 7:09am.

Equation of line:

$y - y_1 = m(x - x_1)$

$y + 2 = \dfrac{4 + 2}{8.5 - 2.5}(x - 2.5)$

$x = y + 4.5$   ←   $x_R$

 

img_0056.jpg

 

Equation of parabola:

$y^2 = 4a(x - h)$

$y^2 = 2(x - 0.5)$

$x = \frac{1}{2}y^2 + 1$   ←   $x_L$

 

$\displaystyle A = \int_{y_1}^{y_2} (x_R - x_L) \, dy$

$\displaystyle A = \int_{-2}^{4} \left[ (y + 4.5) - \left( \frac{1}{2}y^2 + 1\right) \right] \, dy$

Leamir M. Sakal's picture

Submitted by Leamir M. Sakal on August 28, 2017 - 11:16am.

Thanks men! Love the approach