Use a graph of $\displaystyle \int^2_0 \sin 2 \pi x \cos 5 \pi x dx$ to guess the value of the integral. Then prove that your guess is correct by using Calculus.

Based from the graph, it seems that the average value of the area is equal to 0 since the curve above the $x$-axis is the same as the curve below will cancel out since they have opposite magnitude.

Now, by using Calculus,

$ \begin{equation} \begin{aligned} \sin (7 \pi x) &= \sin (2 \pi x + 5 \pi x) = \sin (2 \pi x \cos (5 \pi x) + \cos (2 \pi x) \sin (5 \pi x))\\ \\ & \text{and}\\ \\ \sin (3 \pi x) &= \sin ( 5 \pi x - 2 \pi x) = \sin (5 \pi x ) \cos ( 2 \pi x) - \cos (5 \pi x ) \sin (2 \pi x) \end{aligned} \end{equation} $

So,

$\sin (7 \pi x) - \sin (3 \pi x) = \sin (2 \pi x \cos (5 \pi x) + \cos (2 \pi x) \sin (5 \pi x)) - \sin (5 \pi x ) \cos ( 2 \pi x) - \cos (5 \pi x ) \sin (2 \pi x) = 2 \sin (2 \pi x) \cos (5 \pi x)$

Thus,

$\displaystyle \sin (2 \pi x) \cos (5\pi x) = \frac{\sin (7 \pi x ) - \sin (3 \pi x)}{2}$

Hence,

$ \begin{equation} \begin{aligned} \int^2_0 \sin (2 \pi x) \cos (5 \pi x) dx &= \int^2_0 \left[ \frac{\sin(7 \pi x) - \sin (3 \pi x)}{2} \right] dx\\ \\ &= \frac{1}{2} \left[ \frac{\cos (7\pi x)}{7 \pi} - \frac{\cos (3 \pi x)}{3 \pi} \right]^2_0 \\ \\ &= 0 \end{aligned} \end{equation} $

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