Question

if $f(x)=\int_{x}^{x^{4}} t^{2}dt$ then $f(x)=$

Explanation:

Step1: Recall the Fundamental Theorem of Calculus Part 1, adjusted for variable bounds.

For $F(x)=\int_{a(x)}^{b(x)} g(t)dt$, $F'(x)=g(b(x))\cdot b'(x)-g(a(x))\cdot a'(x)$

Step2: Identify $g(t)$, $a(x)$, $b(x)$

$g(t)=t^2$, $a(x)=x$, $b(x)=x^4$

Step3: Compute derivatives of bounds

$a'(x)=1$, $b'(x)=4x^3$

Step4: Substitute into the formula

$g(b(x))=(x^4)^2=x^8$, $g(a(x))=x^2$
$f'(x)=x^8\cdot4x^3 - x^2\cdot1$

Step5: Simplify the expression

$f'(x)=4x^{11}-x^2$

Answer:

$4x^{11}-x^2$