integral t^2dt from x to x^2
f'(x)=?
well, you can integrate and then take the derivative, and get
f(x) = 1/3 (x^6-x^3)
f'(x) = 2x^5 - x^2
or you can use Leibnitz' Rule and get
(x^2)^2(2x) - (x)^2(1)
= 2x^5 - x^2
read up on it at
http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign
To find \(f'(x)\) when \(f(x) = \int t^2 \, dt\) from \(x\) to \(x^2\), we need to evaluate the integral first.
The integral expression can be rewritten as:
\(f(x) = \int_{x}^{x^2} t^2 \, dt\)
To evaluate this integral, we can use the power rule of integration, which states that the integral of \(t^n\) with respect to \(t\) is \(\frac{{t^{n+1}}}{{n+1}}\) plus a constant.
Applying the power rule, we get:
\(f(x) = \frac{{t^3}}{3} + C\)
Where \(C\) is the constant of integration. Now, we can proceed to find \(f'(x)\) by differentiating \(f(x)\) with respect to \(x\).
\(f'(x) = \frac{{d}}{{dx}} \left( \frac{{t^3}}{3} + C \right)\)
Since the constant of integration does not depend on \(x\), its derivative will be zero. Therefore, the derivative of \(f(x)\) will be:
\(f'(x) = \frac{{d}}{{dx}} \left( \frac{{t^3}}{3} \right)\)
Applying the power rule of differentiation, we bring down the exponent and reduce it by 1:
\(f'(x) = \frac{{1}}{{3}} \cdot 3t^2\)
Simplifying further:
\(f'(x) = t^2\)
So, \(f'(x) = t^2\) when \(f(x) = \int t^2 \, dt\) from \(x\) to \(x^2\).