Question

question 4
if $f(x)=\int_{x}^{x^3} t^2 dt$
then
$f(x) = $
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question 5

Explanation:

Step1: Recall the Fundamental Theorem of Calculus

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

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

Here, $g(t)=t^2$, $a(x)=x$, $b(x)=x^3$

Step3: Compute derivatives of bounds

$a'(x) = \frac{d}{dx}(x) = 1$, $b'(x) = \frac{d}{dx}(x^3) = 3x^2$

Step4: Apply the theorem

$f'(x) = (x^3)^2 \cdot 3x^2 - x^2 \cdot 1$
Simplify: $(x^6)\cdot3x^2 - x^2 = 3x^8 - x^2$

Answer:

$3x^8 - x^2$