Evaluate the improper iterated integral \int\limits_0^3 \int\limits_x^{3} e^{ y^2}...


Evaluate the improper iterated integral

{eq}\int\limits_0^3 \int\limits_x^{3}-e^{-y^2} \text{d}x\text{d} y {/eq}

Double Integrals:

The integral of two orders are solved with either the same order or by reversing the integral order. If the order is to be reversed then the integral will be having the integral limits also changed, but not the final result.

Answer and Explanation:

The double integral given in the question has en error, the correct double integral is:

{eq}\displaystyle \int _0^3\int _y^3-e^{-y^2}dxdy {/eq}

Now we will now solve the inner part and then the outer part, as follows:

{eq}\displaystyle \Rightarrow \int _y^3-e^{-y^2}dx\\ \displaystyle =\left[\left(-e^{-y^2}\right)x\right]^3_y\\ \displaystyle =-3e^{-y^2}+ye^{-y^2}\\ {/eq}

So now the integral of order dy is integrated as follows:

{eq}\displaystyle \Rightarrow \int _0^3-3e^{-y^2}+ye^{-y^2}dy\\ \displaystyle =-\int _0^33e^{-y^2}dy+\int _0^3ye^{-y^2}dy\\ \displaystyle =3\left[\frac{\sqrt{\pi }}{2}\text{erf}\left(y\right)\right]^3_0 +\frac{1}{2}\left[e^{-y^2}\right]_0^3 ~~~~~~~~~~~~~~~~~~\left [ \because \int \:e^{-y^2}dy=\frac{\sqrt{\pi }}{2}\text{erf}\left(y\right) \right ]\\ \displaystyle =-\frac{3\left(\sqrt{\pi }\text{erf}\left(3\right)-\sqrt{\pi }\text{erf}\left(0\right)\right)}{2}+\frac{e^9-1}{2e^9} {/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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