Question

fq(x) = integrate sqrt(4 + z ^ 6) dz from 0 to x ^ s * then

Answer 1

We are given the definite integral:

$q(x)=\int_0^{x^8}\sqrt{4+z^6}dz$

We need to find q'(x).

We can use the fundamental theorem of calculus to solve this. Given a function F(x):

$F(x)=\int_a^bf(x)dx$

Then:

$F^{\prime}(x)=f(x)$

Thus, we can find the antiderivative q'(x), using the fundamental theorem of calculus and the chain rule:

$\frac{d}{dx}(\int_0^{x^8}\sqrt{4+z^6}dz)=\sqrt{4+(x^8)^6}\cdot\frac{d}{dx}(x^8)=\sqrt{4+x^{48}}\cdot8x^7$

The answer is:

$q^{\prime}(x)=8x^7\sqrt{4+x^{48}}$