Evaluate the integral. \int_{-3}^{2} \sin(\pi t)dt

Question:

Evaluate the integral.

{eq}\int_{-3}^{2} \sin(\pi t)dt {/eq}

Integrations Using Substitutions

First we assume u is equal to some value in order to simplify the solution

As we are assuming u is equal to some value limits also changes accordingly

Then we integrate the function and substitute the limits

Formulas Used

{eq}\displaystyle \begin{align} \int a\cdot f\left(x\right)dx&=a\cdot \int f\left(x\right)dx\\ \int \sin \left(u\right)du&=-\cos \left(u\right)\\ \end{align} {/eq}

Answer and Explanation:

Given

{eq}\displaystyle \int_{-3}^{2} \sin(\pi t)dt {/eq}

Assume

{eq}\displaystyle \begin{align} u&=\pi t\\ du&=\pi dt\\ dt&=\frac{du}{\pi}\\ \end{align} {/eq}

Therefore

{eq}\displaystyle \begin{align} \int_{-3}^{2} \sin(\pi t)dt&=\int _{-3\pi }^{\pi 2}\sin \left(u\right)\frac{1}{\pi }du\\ &=\frac{1}{\pi }\cdot \int _{-3\pi }^{\pi 2}\sin \left(u\right)du\\ &=\frac{1}{\pi }\left[-\cos \left(u\right)\right]^{\pi 2}_{-3\pi }\\ &=\frac{1}{\pi }\left(-\cos(2\pi)+\cos(-3\pi)\right)\\ &=\frac{1}{\pi }\left(-1+(-1)\right)\\ &=\frac{1}{\pi }\left(-1-1\right)\\ &=\frac{-2}{\pi } \end{align} {/eq}

Therefore the final result is

{eq}\displaystyle \color{blue}{\int_{-3}^{2} \sin(\pi t)dt=\frac{-2}{\pi}} {/eq}


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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