I was working on a problem $\int^1_0$8sin(5$\pi$x)dx earlier and was shown the solution on the site I'm using to complete this work. I'd like an explanation on how, at a later point in my process/the way the site shows it, $8\over3\pi$$\int^1_0sin(u)du$ becomes $16\over3\pi$. I calculated that the upper limit was 3$\pi$ and the lower limit was 0. What do I plug in to get me$16\over3\pi$ since if I put my upper and lower limit into my equation, I get 0-0?
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The antiderivative of $\sin$ is $-\cos$ – Weston Miller May 01 '20 at 01:08
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$-\cos(3\pi) = 1$ and $-\cos(0) = -1$ so $1--1 = 2$ – Weston Miller May 01 '20 at 01:09
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Wish the step-by-step solution I was following told me to take the antiderivative... thanks for the help!! – Rachel May 01 '20 at 01:14
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$\sin5\pi x$ or $\sin3\pi x$? – Koro May 01 '20 at 01:16