# Evaluate the double integral \int_1^e \int_0^{3\ln x} 5x \sin y \ dy \ dx

## Question:

Evaluate the double integral {eq}\int_1^e \int_0^{3\ln x} 5x \sin y \ dy \ dx {/eq}

## Double Integrals:

The double integral is solved with the trapezoidal rule if the outer part with the definite limits of integral is to be approximated. The result of the double integral will be finite in this case.

## Answer and Explanation:

The double integral give is:

{eq}\displaystyle \int_1^e \int_0^{3\ln x} 5x \sin y \ dy \ dx {/eq}

We will first solve the definite integral:

{eq}\displaystyle \int _0^{3ln\:x}\:5x\:sin\:y\:dy\\ \displaystyle =5x\left[-\cos \left(y\right)\right]^{3\ln \left(x\right)}_0\\ \displaystyle =5x\left(-\cos \left(3\ln \left(x\right)\right)+1\right)\\ {/eq}

Now the outer integral is;

{eq}\displaystyle \int _1^e5x\left(-\cos \:\left(3\ln \:\left(x\right)\right)+1\right)\:dx {/eq}

This will be solved by the trapezoidal rule as we can approximate this integral:

{eq}\int_{a}^{b}f(x)dx= \frac{f(a)+f(b)}{2}(b-a) \\ \displaystyle \Rightarrow \int _1^e5x\left(-\cos \:\left(3\ln \:\left(x\right)\right)+1\right)\:dx\\ \displaystyle =\frac{5e\left(-\cos \left(3\ln \left(e\right)\right)+1\right)+5\left(-\cos \left(3\ln \left(1\right)\right)+1\right)}{2}\left(e-1\right)\\ =23.23\\ {/eq}

So the double integral value is:

23.23

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from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14