Is it just me.... (1 Viewer)

Run hard@thehsc · Apr 16, 2022

or did they integrate this wrong?

Also tell me what you guys get for this:

The answers seem uncorrected lol

5uckerberg · Apr 16, 2022

The first one is all right but yeah some working errors. At best this should be one less than full marks, at worst two less than full marks for this question.

The second Q is pretty straightforward. Trig expansion is your ally $\int_{0}^{\frac{\pi}{3}}\sin{5x}\sin{2x}dx$ . This I guarantee should give you $\frac{1}{2}\int_{0}^{\frac{\pi}{3}}\left(\cos{3x}-\cos{7x}\right)dx$

In step 2 we will integrate the integral.
$\frac{1}{2}\left[\frac{1}{3}\sin{3x}-\frac{1}{7}\sin{7x}\right]_{0}^{\frac{\pi}{3}}$

$\frac{1}{2}\left[\frac{1}{3}\times{\sin{\pi}}-\frac{1}{7}\times{\sin{\frac{7\pi}{3}}}\right]$
$-\frac{1}{14}\times{\frac{\sqrt{3}}{2}}$
$\frac{-\sqrt{3}}{28}$

Run hard@thehsc · Apr 17, 2022

@5uckerberg yep I got that answer too - for some reason the answers published forgot the minus.... Thanks

Is it just me.... (1 Viewer)

Run hard@thehsc

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5uckerberg

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Run hard@thehsc

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