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I was testing my calculus knowledge when I found an example final exam from UCIrvine: http://www.math.uci.edu/sites/math.uci.edu/files/2B_final_samp1.pdf

Number 2.) a.) asks to evaluate:

$$ {d\over dx}\int_{sin(x)}^{x^2}t^3tan(t)dt $$

This is what I did:

$\begin{aligned} {d\over dx}\int_{\sin(x)}^{x^2}t^3\tan(t)dt &={d\over dx}\left(\int_{\sin(x)}^{a}t^3\tan(t)dt+\int_{a}^{x^2}t^3\tan(t)dt\right) \\ &= {d\over dx}\left(-\int_{a}^{\sin(x)}t^3\tan(t)dt+\int_{a}^{x^2}t^3\tan(t)dt\right) \\ &= -{d\over dx}\int_{a}^{\sin(x)}t^3\tan(t)dt+{d\over dx}\int_{a}^{x^2}t^3\tan(t)dt \\ &= -\sin^3(x)\tan(\sin(x))\cos(x)+{x^2}^3\tan(x^2)2x \\ &= 2x^7\tan(x^2)-\sin^3(x)\cos(x)\tan(\sin(x)) \end{aligned}$

Is this correct?

blue_egg
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kukac67
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2 Answers2

23

Yes is correct, remember that $$\frac{d}{dx}\int_{g(x)}^{f(x)}h(t)\,dt=h(f(x))\cdot f'(x)-h(g(x))\cdot g'(x) $$

this is by the second theorem of calculus and by chain rule.

AsdrubalBeltran
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1

You don't need to break up the integral. You can start from the first term $\frac{d}{dx} \int_{\sin(x)}^{x^2} t^3 \tan(t) dt$. Leibniz's rule gives a direct formula for this derivative.

The other answer basically says
$\begin{aligned} \frac{d}{dt} \left[\int_{a(x)}^{b(x)} f(t) dt\right] &= \frac{d}{dt} \left[F(a(x)) - F(b(x))\right] \;\; \text{ by FTC} \\ &= f(a(x))a'(x) - f(b(x))b'(x) \;\; \text{ by chain rule} \end{aligned}$

where $F$ is the antiderivative of $f$. More generally you can use something like this derivation to get the more general form when $f = f(x,t)$.

blue_egg
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