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The below figure shows the graph of $y=f(x)$ for $-4\leq x \leq4. $

y=f(x)

and $g(x)=\int_{-4}^xf(t)$dt for $-4\leq x \leq4. $

There are 2 questions that I'm not able to solve.
1) At what value of $x$ is $g(x)$ minimum

2) how to sketch the graph of $y=g(x)$

I'm not able find a way to solve above questions, Please assist. Also any resource I should refer Please let me know.

Thanks, Arif

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    Do you know what the geometric interpretation of the integral $\int_{-4}^xf(t)\,\mbox{d}t$ is? – 2017-02-23
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    For (1), you can differentiate g(x) and use normal max/min arguments. Look up "differentiating an integral" maybe. – 2017-02-23
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    @StackTD I understand $\int_{-4}^xf(t)$ dt. means area of the region bounded by $f(t)~~ x$ axis and the two limits. Also we don't have function $g(x)$ for differentiation then how should be go about – 2017-02-25
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    I understand we'll use the Fundamental Theorem of calculus that will give derivative of $g(x)$ as $f(x)$,since derivative of $g(x)=\int_{-a}^xf(t)$dt is $f(x)$, and then g\`(x) is $0$ at $x=1$, hence answer for first part, please let me know if that's correct. Also I still need to figure out how to do question 2 ? – 2017-02-25

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