f(x) = int_{cos x}^{sin x} (1 - t + 2t^3) dt | vedclass

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(B) Using Leibniz's rule for differentiation under the integral sign: $f'(x) = (1 - \sin x + 2\sin^3 x)(\cos x) - (1 - \cos x + 2\cos^3 x)(-\sin x)$ $

Vedclass · Accepted Answer

a maximum at $\frac{3\pi}{4}$ and a minimum at $\frac{7\pi}{4}$

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(B) Using Leibniz's rule for differentiation under the integral sign: $f'(x) = (1 - \sin x + 2\sin^3 x)(\cos x) - (1 - \cos x + 2\cos^3 x)(-\sin x)$ $f'(x) = \cos x - \sin x \cos x + 2\sin^3 x \cos x + \sin x - \sin x \cos x + 2\cos^3 x \sin x$ $f'(x) = \cos x + \sin x - 2\sin x \cos x + 2\sin x \cos x (\sin^2 x + \cos^2 x)$ $f'(x) = \cos x + \sin x - 2\sin x \cos x + 2\sin x \cos x = \cos x + \sin x$ For critical points,set $f'(x) = 0$: $\cos x + \sin x = 0 \implies 	an x = -1$ In $[0, 2\pi]$,$x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$ $f''(x) = -\sin x + \cos x$ At $x = \frac{3\pi}{4}$,$f''(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} At $x = \frac{7\pi}{4}$,$f''(\frac{7\pi}{4}) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \sqrt{2} > 0$ (Minimum) Thus,the function has a maximum at $\frac{3\pi}{4}$ and a minimum at $\frac{7\pi}{4}$.

$f(x) = \int_{\cos x}^{\sin x} (1 - t + 2t^3) dt$ has in $[0, 2\pi]$:

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