Find the derivative of f(x) = ∫_1^{x^2​} (t^2 + t + 1) dt

[hndalama]

Find the derivative of [FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Size1]∫^{x^2}_​1[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]t²[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT]. to be clear this is integral of the integrand with upperendpoint = x² and lower endpoint=​1
I don't know what to do with the x². The answer given is 2x(x²​ + x + 1) but i don't know how they got this.

 

[pka]

Find the derivative of [FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Size1]∫^{x^2}_​1[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]t²[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT]. to be clear this is integral of the integrand with upperendpoint = x² and lower endpoint=​1. I don't know what to do with the x². The answer given is 2x(x²​ + x + 1) but i don't know how they got this.

Assuming that each of \(\displaystyle f~\&~g\) is a differential function then:
\(\displaystyle \displaystyle{\dfrac{d}{{dx}}\left[ {\int_{g(x)}^{f(x)} {\Phi (t)dt} } \right] = \Phi (f(x))f'(x) - \Phi (g(x))g'(x)}\)

 

[Steven G]

Find the derivative of [FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Size1]∫^{x^2}_​1[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]t²[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT]. to be clear this is integral of the integrand with upperendpoint = x² and lower endpoint=​1
I don't know what to do with the x². The answer given is 2x(x²​ + x + 1) but i don't know how they got this.

pka showed the way of doing this--did you not know this formula?

Learn that formula!!

Having said all that why did you not compute that simple integral, evaluate it from 0 to x² which will give you a function of x and then differentiate that function?

 

[hndalama]

when I use the formula pka gave, the answer I keep getting is 2x(x⁴ + x² + 1) which is different to the given answer 2x(x² + x + 1).

when I evaluate it in the way that jomo described I still get 2x(x⁴ + x² + 1)

is the answer wrong?

 

[HallsofIvy]

It is not at all difficult to do this "directly":
\(\displaystyle \int_1^{x^2} (t^2+ t+ 1) dt= \left[\frac{t^3}{3}+ \frac{t^2}{2}+ t\right]_1^{x^2}= \frac{x^6}{3}+ \frac{x^4}{2}+ x^2- \frac{1}{3}- \frac{1}{2}- 1= \frac{x^6}{3}+ \frac{x^4}{2}+ x^2- \frac{11}{6}\)
and the derivative of that is \(\displaystyle 2x^5+ 2x^3+ 2x\).

But we can also write \(\displaystyle u= x^2\) so that the integral is \(\displaystyle F(u)= \int_1^u (t^2+ t+ 1)dt\) and, by the chain rule, \(\displaystyle \frac{dF}{dx}= \frac{dF}{du}\frac{du}{dx}\).

By the "fundamental theorem of Calculus", \(\displaystyle \frac{dF}{du}= u^2+ u+ 1\) and, or course, \(\displaystyle \frac{du}{dx}= \frac{dx^2}{dx}= 2x\) so \(\displaystyle \frac{dF}{du}\frac{du}{dx}= (u^2+ u+ 1)(2x)= (x^4+ x^2+ 1)(2x)= 2x^5+ 2x^3+ 2x\) as above.

The "given answer" is incorrect.

 

[hndalama]

It is not at all difficult to do this "directly":
\(\displaystyle \int_1^{x^2} (t^2+ t+ 1) dt= \left[\frac{t^3}{3}+ \frac{t^2}{2}+ t\right]_1^{x^2}= \frac{x^6}{3}+ \frac{x^4}{2}+ x^2- \frac{1}{3}- \frac{1}{2}- 1= \frac{x^6}{3}+ \frac{x^4}{2}+ x^2- \frac{11}{6}\)
and the derivative of that is \(\displaystyle 2x^5+ 2x^3+ 2x\).

But we can also write \(\displaystyle u= x^2\) so that the integral is \(\displaystyle F(u)= \int_1^u (t^2+ t+ 1)dt\) and, by the chain rule, \(\displaystyle \frac{dF}{dx}= \frac{dF}{du}\frac{du}{dx}\).

By the "fundamental theorem of Calculus", \(\displaystyle \frac{dF}{du}= u^2+ u+ 1\) and, or course, \(\displaystyle \frac{du}{dx}= \frac{dx^2}{dx}= 2x\) so \(\displaystyle \frac{dF}{du}\frac{du}{dx}= (u^2+ u+ 1)(2x)= (x^4+ x^2+ 1)(2x)= 2x^5+ 2x^3+ 2x\) as above.

The "given answer" is incorrect.

thank you