Evaluate the integral. \int_{0}^{6} \int_{0}^{2} (10x^{2}y + 8xy) dy dx


Evaluate the integral.

{eq}\displaystyle\; \int_{0}^{6} \int_{0}^{2} \left(10x^{2}y + 8xy\right) dy\,dx {/eq}

Double Integration:

The integral that has both upper and the lower limits of integration is known as a definite integral.

Double integration is a type of integration in which the integral involves the two variables {eq}(x \ and \ y) {/eq} and we have to integrate with respect to both the variable one by one.

Answer and Explanation:

We have to evaluate the integral

{eq}\displaystyle\; \int_{0}^{6} \int_{0}^{2} \left(10x^{2}y + 8xy\right) dy\,dx {/eq}

First, we integrate the integral with respect to {eq}y {/eq} and then integrate with respect to {eq}x {/eq}.

{eq}\begin{align} \int_{0}^{6} \int_{0}^{2} (10x^{2}y + 8xy) dy dx &=\int_{0}^{6} \left (\int_{0}^{2} (10x^{2}y + 8xy) dy \right ) dx\\ &=\int_{0}^{6} \left (\int _0^210x^2ydy+\int _0^28xydy \right )dx \ & \left [ \because \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx \right ]\\ &=\int_{0}^{6}\left ( 10x^2\cdot \int _0^2ydy+8x\cdot \int _0^2ydy \right )dx \ & \left [ \because \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx \right ]\\ &=\int_{0}^{6}\left ( 10x^2\left ( \dfrac{y^2}{2} \right )_0^2+8x\left ( \dfrac{y^2}{2} \right )_0^2 \right )dx \ & \left [ \int x^adx=\dfrac{x^{a+1}}{a+1} \right ]\\ &=\int_{0}^{6}\left (5x^2(y^2)_0^2+4x(y^2)_0^2 \right )dx\\ &=\int_{0}^{6}\left (\left ( 5x^2(2^2-0^2) \right )+4x(2^2-0^2) \right )dx\\ &=\int_{0}^{6}\left ( 5x^2(4)+4x(4) \right )dx\\ &=\int_{0}^{6}\left ( 20x^2+16x \right )dx\\ &=\int_{0}^{6}\left ( 20x^2+16x \right )dx\\ &=\int_{0}^{6} 20x^2dx+\int_{0}^{6}16xdx\ & \left [\because \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx \right ]\\ &=20\int_{0}^{6}x^2dx+16\int_{0}^{6}xdx\ & \left [ \because \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx \right ]\\ &=20\left ( \dfrac{x^{2+1}}{2+1} \right )_{0}^{6}+16\left ( \dfrac{x^{1+1}}{1+1} \right )_{0}^{6}\ & \left [ \because \int x^adx=\dfrac{x^{a+1}}{a+1} \right ]\\ &=20\left ( \dfrac{x^3}{3} \right )_{0}^{6}+16\left ( \dfrac{x^2}{2} \right )_{0}^{6}\\ &=20\left ( \dfrac{6^3}{3}-0 \right )+16\left ( \dfrac{6^2}{2}-0 \right )\\ &=20\left ( \dfrac{216}{3} \right )+16\left ( \dfrac{36}{2} \right )\\ &=20(72)+16(18)\\ &=1440+288\\ &=1728 \end{align} {/eq}

{eq}\color{blue}{\boxed{\int_{0}^{6} \int_{0}^{2} (10x^{2}y + 8xy) dy dx=1728}} {/eq}

Learn more about this topic:

Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14

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