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:

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

Is this correct?


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.


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