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• According to what is given,
[tex]\mathsf{g(x)=\displaystyle\int_0^x sin(t^2)\,dt\qquad\qquad(-1\le x\le 3)}[/tex]
• Now, differentiate g by using the Fundamental Theorem of Calculus:
[tex]\mathsf{\displaystyle g'(x)=\frac{d}{dx}\int_0^x sin(t^2)\,dt}\\\\\\ \mathsf{\displaystyle g'(x)=sin(x^2)}[/tex]
• g is increasing in the interval where g'(x) is positive. So now, just solve this inequality:
[tex]\mathsf{g'(x)\ \textgreater \ 0}\\\\ \mathsf{sin(x^2)\ \textgreater \ 0}[/tex]
• The sine function is positive for angles that lie either in the first or the second quadrant. So,
[tex]\mathsf{0\ \textless \ x^2\ \textless \ \pi}[/tex]
• The inequality above involves only non-negative terms. So, the sign of the inequality keeps the same for the square root of those terms:
[tex]\mathsf{0\ \textless \ x\ \textless \ \sqrt{\pi}\qquad\quad(i)}[/tex]
• Checking the intersection between the interval we just found above and the domain of g:
Notice that
[tex]-1\le 0\ \textless \ x\ \textless \ \sqrt{\pi}\le 3[/tex]
which implies that
[tex]\mathsf{\left]0,\,\sqrt{\pi}\right[\subset [1,\,3]}\\\\ \mathsf{\left]0,\,\sqrt{\pi}\right[\subset Dom(g)}.[/tex]
Therefore,
g is increasing on the interval [tex]\mathsf{\left]0,\,\sqrt{\pi}\right[.}[/tex]
I hope this helps. =)
Tags: derivative fundamental theorem of calculus increasing interval differential integral calculus
_______________
• According to what is given,
[tex]\mathsf{g(x)=\displaystyle\int_0^x sin(t^2)\,dt\qquad\qquad(-1\le x\le 3)}[/tex]
• Now, differentiate g by using the Fundamental Theorem of Calculus:
[tex]\mathsf{\displaystyle g'(x)=\frac{d}{dx}\int_0^x sin(t^2)\,dt}\\\\\\ \mathsf{\displaystyle g'(x)=sin(x^2)}[/tex]
• g is increasing in the interval where g'(x) is positive. So now, just solve this inequality:
[tex]\mathsf{g'(x)\ \textgreater \ 0}\\\\ \mathsf{sin(x^2)\ \textgreater \ 0}[/tex]
• The sine function is positive for angles that lie either in the first or the second quadrant. So,
[tex]\mathsf{0\ \textless \ x^2\ \textless \ \pi}[/tex]
• The inequality above involves only non-negative terms. So, the sign of the inequality keeps the same for the square root of those terms:
[tex]\mathsf{0\ \textless \ x\ \textless \ \sqrt{\pi}\qquad\quad(i)}[/tex]
• Checking the intersection between the interval we just found above and the domain of g:
Notice that
[tex]-1\le 0\ \textless \ x\ \textless \ \sqrt{\pi}\le 3[/tex]
which implies that
[tex]\mathsf{\left]0,\,\sqrt{\pi}\right[\subset [1,\,3]}\\\\ \mathsf{\left]0,\,\sqrt{\pi}\right[\subset Dom(g)}.[/tex]
Therefore,
g is increasing on the interval [tex]\mathsf{\left]0,\,\sqrt{\pi}\right[.}[/tex]
I hope this helps. =)
Tags: derivative fundamental theorem of calculus increasing interval differential integral calculus
The derivative fundamental theorem of the calculus increasing interval differential integral calculus.
What is a function?
The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.
Let g be the function given by
[tex]g(x) = \int_ 0 ^ x \sin t^2 dt \ \ \ \ -1 \leq x \leq 3[/tex]
Now, differentiate g(x), then we have
[tex]g'(x) = \sin x^2[/tex]
g(x) is increasing in the interval where g'(x) is positive. Then we have
g'(x) > 0
sin (x²) > 0
We know that
0 < x² < π
0 < x < √π
Checking the intersection between the interval, we just found above and the domain of g(x).
[tex]-1 \leq 0 < x < \sqrt{\pi} \leq 3[/tex]
Which implies that
[tex]]0, \sqrt{\pi}[ \subset [1, 3][/tex]
Therefore, g is increasing on the interval [0, √π]
More about the function link is given below.
https://brainly.com/question/5245372
