Integral calculus:Application Member for 10 years 6 months Submitted by Leamir M. Sakal on Tue, 05/09/2017 - 22:48 Problem: Find the area of the region bounded by the graphs as shown in the picture, given that the 4a=2. Tags Area by integration Integral application Log in or register to post comments Equation of line: Member for 17 years 6 months Jhun Vert Wed, 06/07/2017 - 07:09 Equation of line: $y - y_1 = m(x - x_1)$ $y + 2 = \dfrac{4 + 2}{8.5 - 2.5}(x - 2.5)$ $x = y + 4.5$ ← $x_R$ Equation of parabola: $y^2 = 4a(x - h)$ $y^2 = 2(x - 0.5)$ $x = \frac{1}{2}y^2 + 1$ ← $x_L$ $\displaystyle A = \int_{y_1}^{y_2} (x_R - x_L) \, dy$ $\displaystyle A = \int_{-2}^{4} \left[ (y + 4.5) - \left( \frac{1}{2}y^2 + 1\right) \right] \, dy$ Log in or register to post comments Thanks men! Love the approach Member for 10 years 6 months Leamir M. Sakal Mon, 08/28/2017 - 11:16 In reply to Equation of line: by Jhun VertThanks men! Love the approach Log in or register to post comments
Equation of line: Member for 17 years 6 months Jhun Vert Wed, 06/07/2017 - 07:09 Equation of line: $y - y_1 = m(x - x_1)$ $y + 2 = \dfrac{4 + 2}{8.5 - 2.5}(x - 2.5)$ $x = y + 4.5$ ← $x_R$ Equation of parabola: $y^2 = 4a(x - h)$ $y^2 = 2(x - 0.5)$ $x = \frac{1}{2}y^2 + 1$ ← $x_L$ $\displaystyle A = \int_{y_1}^{y_2} (x_R - x_L) \, dy$ $\displaystyle A = \int_{-2}^{4} \left[ (y + 4.5) - \left( \frac{1}{2}y^2 + 1\right) \right] \, dy$ Log in or register to post comments
Thanks men! Love the approach Member for 10 years 6 months Leamir M. Sakal Mon, 08/28/2017 - 11:16 In reply to Equation of line: by Jhun VertThanks men! Love the approach Log in or register to post comments
Equation of line:
Member for
17 years 6 monthsEquation of line:
$y + 2 = \dfrac{4 + 2}{8.5 - 2.5}(x - 2.5)$
$x = y + 4.5$ ← $x_R$
Equation of parabola:
$y^2 = 2(x - 0.5)$
$x = \frac{1}{2}y^2 + 1$ ← $x_L$
$\displaystyle A = \int_{y_1}^{y_2} (x_R - x_L) \, dy$
$\displaystyle A = \int_{-2}^{4} \left[ (y + 4.5) - \left( \frac{1}{2}y^2 + 1\right) \right] \, dy$
Thanks men! Love the approach
Member for
10 years 6 monthsIn reply to Equation of line: by Jhun Vert
Thanks men! Love the approach