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Korvikt [17]
3 years ago
6

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

Mathematics
1 answer:
Amanda [17]3 years ago
4 0

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

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