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(\mathrm d)/(\mathrm dx)\int_1^(\cos x)(t+\sqrt t)\,\mathrm dt=(\cos x+√(\cos x))(\mathrm d)/(\mathrm dx)\cos x=-\sin x(\cos x+√(\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=\frac{t^2}2+\frac{2t^(3/2)}3\bigg|_(t=1)^(t=\cos x)=\left(\frac{\cos^2x}2+\frac{2(\cos x)^(3/2)}3\right)-\left(\frac12+\frac23\right)`

`=\frac{\cos^2x}2+\frac{2√(\cos^3x)}3-\frac76`

c. The derivative of the previous result is

`\frac{2\cos x(-\sin x)}2+\frac{2\cdot\frac32(\cos x)^(1/2)(-\sin x)}3=-\sin x\cos x-\sin x√(\cos x)`

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

d. ... yes