Question

CALCULUS EXPERT WANTED. Can someone solve this or at least try to explain to me the fundamental theorem of calculus (FTC).

Answer 1

a. By the FTC,

$\displaystyle\frac{\mathrm d}{\mathrm dx}\int_1^{\cos x}(t+\sqrt t)\,\mathrm dt=(\cos x+\sqrt{\cos x})\dfrac{\mathrm d}{\mathrm dx}\cos x=-\sin x(\cos x+\sqrt{\cos x})$

b. We can either evaluate the integral directly, or take the integral of the previous result. With the first method, we get

$\displaystyle\int_1^{\cos x}(t+\sqrt t)\,\mathrm dt=\dfrac{t^2}2+\dfrac{2t^{3/2}}3\bigg|_{t=1}^{t=\cos x}=\left(\dfrac{\cos^2x}2+\dfrac{2(\cos x)^{3/2}}3\right)-\left(\dfrac12+\dfrac23\right)$

$=\dfrac{\cos^2x}2+\dfrac{2\sqrt{\cos^3x}}3-\dfrac76$

c. The derivative of the previous result is

$\dfrac{2\cos x(-\sin x)}2+\dfrac{2\cdot\frac32(\cos x)^{1/2}(-\sin x)}3=-\sin x\cos x-\sin x\sqrt{\cos x}$

which is the same as the answer given in part (a), so ...

d. ... yes