Evaluate the double integral \int_{-1}^1 \int_{y^2}^y (2x -y) \, dx \, dy

Question:

Evaluate the double integral

{eq}\int_{-1}^1 \int_{y^2}^y (2x -y) \, dx \, dy {/eq}

Double Integration:

Double integral is a definite integral which involves two variable.This is used to find the two dimensional area of the given problem. In this problem, we will find the double integration first with respect to x and then with respect to y.

Answer and Explanation:

Given that {eq}\int_{-1}^1 \int_{y^2}^y (2x -y) \, dx \, dy {/eq}

First integrate with respect to x and then integrate with respect to y.

{eq}\int_{-1}^1 \int_{y^2}^y (2x -y) \, dx \, dy\\ =\int_{-1}^1 \left [-\int _{y^2}^yydx+\int _{y^2}^y2xdx \right ] \, dy\\ =\int_{-1}^1 \left [ \left[yx\right]^y_{y^2}+2\left[\frac{x^2}{2}\right]^y_{y^2} \right ] \, dy\\ =\int_{-1}^1\left (-\left(y^2-y^3\right)+y^2-y^4 \right )dy\\ =\int _{-1}^1\left(-y^4+y^3\right)dy\\ =-\int _{-1}^1y^4dy+\int _{-1}^1y^3dy\\ =-\left[\frac{y^5}{5}\right]^1_{-1}+\left[\frac{y^4}{4}\right]^1_{-1}\\ =-\frac{2}{5}+0\\ =-\frac{2}{5} {/eq}


Learn more about this topic:

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Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
498

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