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_______________
• According to what is given,
• Now, differentiate g by using the Fundamental Theorem of Calculus:
<span>• </span>g is increasing in the interval where g'(x) is positive. So now, just solve this inequality:
• The sine function is positive for angles that lie either in the first or the second quadrant. So,
• The inequality above involves only non-negative terms. So, the sign of the inequality keeps the same for the square root of those terms:
• Checking the intersection between the interval we just found above and the domain of g:
Notice that
which implies that
![\mathsf{\left]0,\,\sqrt{\pi}\right[\subset [1,\,3]}\\\\ \mathsf{\left]0,\,\sqrt{\pi}\right[\subset Dom(g)}.](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cleft%5D0%2C%5C%2C%5Csqrt%7B%5Cpi%7D%5Cright%5B%5Csubset%20%5B1%2C%5C%2C3%5D%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cleft%5D0%2C%5C%2C%5Csqrt%7B%5Cpi%7D%5Cright%5B%5Csubset%20Dom%28g%29%7D.)
Therefore,
g is increasing on the interval![\mathsf{\left]0,\,\sqrt{\pi}\right[.}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cleft%5D0%2C%5C%2C%5Csqrt%7B%5Cpi%7D%5Cright%5B.%7D)
I hope this helps. =)
Tags: <em>derivative fundamental theorem of calculus increasing interval differential integral calculus</em>
_______________
• According to what is given,
• Now, differentiate g by using the Fundamental Theorem of Calculus:
<span>• </span>g is increasing in the interval where g'(x) is positive. So now, just solve this inequality:
• The sine function is positive for angles that lie either in the first or the second quadrant. So,
• The inequality above involves only non-negative terms. So, the sign of the inequality keeps the same for the square root of those terms:
• Checking the intersection between the interval we just found above and the domain of g:
Notice that
which implies that
Therefore,
g is increasing on the interval
I hope this helps. =)
Tags: <em>derivative fundamental theorem of calculus increasing interval differential integral calculus</em>
