Question

1. 5 pts, 2.5 each compute the following:

• ( f(2) ), where ( f(x)=int_{0}^{x}left(t^{4}-3 t^{2}

ight) d t )

• ( \frac{d}{d x} int_{x^{2}}^{x^{4}} sqrt{t} d t )

Explanation:

Step1: Apply the Fundamental Theorem of Calculus Part 1

According to the Fundamental Theorem of Calculus Part 1, if \(F(x)=\int_{a}^{x}f(t)dt\), then \(F^{\prime}(x) = f(x)\). For \(F(x)=\int_{0}^{x}(t^{4}-3t^{2})dt\), we have \(F^{\prime}(x)=x^{4}-3x^{2}\).

Step2: Evaluate \(F^{\prime}(x)\) at \(x = 2\)

Substitute \(x = 2\) into \(F^{\prime}(x)\): \(F^{\prime}(2)=2^{4}-3\times2^{2}\).

$$
LATEXBLOCK0
$$

Step3: Apply the Leibniz rule

The Leibniz rule for \(\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)dt=f(v(x))v^{\prime}(x)-f(u(x))u^{\prime}(x)\). Here \(f(t)=\sqrt{t}=t^{\frac{1}{2}}\), \(v(x)=x^{4}\), \(v^{\prime}(x) = 4x^{3}\), \(u(x)=x^{2}\), \(u^{\prime}(x)=2x\).

$$
LATEXBLOCK1
$$

Assuming \(x>0\) (since we are dealing with a typical calculus - level problem where we often consider the principal root in the context of differentiation of integrals without specifying a domain otherwise), \(|x| = x\). Then \(\frac{d}{dx}\int_{x^{2}}^{x^{4}}\sqrt{t}dt=4x^{5}-2x^{2}\)

Answer:

• \(F^{\prime}(2)=4\)
• \(\frac{d}{dx}\int_{x^{2}}^{x^{4}}\sqrt{t}dt = 4x^{5}-2x^{2}\)