Question

Let g be the function given by g(x) = the integral from 0 to x sin(t^2) for -1 < or equal to x < or equal to 3. Find the interval where the function g(x) is increasing.

Answer 1

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•  According to what is given,

$\mathsf{g(x)=\displaystyle\int_0^x sin(t^2)\,dt\qquad\qquad(-1\le x\le 3)}$


•  Now, differentiate g by using the Fundamental Theorem of Calculus:

$\mathsf{\displaystyle g'(x)=\frac{d}{dx}\int_0^x sin(t^2)\,dt}\\\\\\ \mathsf{\displaystyle g'(x)=sin(x^2)}$


•  g is increasing in the interval where g'(x) is positive. So now, just solve this inequality:

$\mathsf{g'(x)\ \textgreater \ 0}\\\\ \mathsf{sin(x^2)\ \textgreater \ 0}$


•  The sine function is positive for angles that lie either in the first or the second quadrant. So,

$\mathsf{0\ \textless \ x^2\ \textless \ \pi}$


•  The inequality above involves only non-negative terms. So, the sign of the inequality keeps the same for the square root of those terms:

$\mathsf{0\ \textless \ x\ \textless \ \sqrt{\pi}\qquad\quad(i)}$


•  Checking the intersection between the interval we just found above and the domain of g:

Notice that

$-1\le 0\ \textless \ x\ \textless \ \sqrt{\pi}\le 3$


which implies that

$\mathsf{\left]0,\,\sqrt{\pi}\right[\subset [1,\,3]}\\\\ \mathsf{\left]0,\,\sqrt{\pi}\right[\subset Dom(g)}.$


Therefore,

g is increasing on the interval  $\mathsf{\left]0,\,\sqrt{\pi}\right[.}$


I hope this helps. =)


Tags:  derivative fundamental theorem of calculus increasing interval differential integral calculus