Question

f(x) = integrate t ^ 4 dt from 0 to x ^ 2 * then

Answer 1

The given function is:

$f(x)=\int_0^{x^2}t^4dt$

Therefore, by evaluating the integral on the right hand side, it follows that:

$\begin{gathered} f(x)=\frac{t^{4+1}}{4+1}\biggr\rvert_0^{x^2} \\ f(x)=\frac{t^5}{5}\biggr\rvert_0^{x^2} \\ f(x)=\frac{(x^2)^5}{5}-\frac{(0)^5}{5} \\ f(x)=\frac{x^{10}}{5} \end{gathered}$

Therefore, by differentiating with respect to x, it follows that:

$\begin{gathered} f^{\prime}(x)=10\times\frac{x^{10-1}}{5} \\ f^{\prime}(x)=2x^9 \end{gathered}$

Therefore,

f'(x) = 2x⁹