Question

if \\( f(x) = \int_{0}^{x^3} t^4 dt \\) then \\( f(x) = \\)

Explanation:

Step1: Recall the Fundamental Theorem of Calculus and Chain Rule

The Fundamental Theorem of Calculus Part 1 states that if \( F(x)=\int_{a}^{x}g(t)dt \), then \( F'(x) = g(x) \). When the upper limit is a function of \( x \), say \( u(x) \), we use the chain rule. So if \( f(x)=\int_{a}^{u(x)}g(t)dt \), then \( f'(x)=g(u(x))\cdot u'(x) \).

Here, \( g(t)=t^{4} \), \( a = 0 \), and \( u(x)=x^{3} \). First, find \( u'(x) \). The derivative of \( u(x)=x^{3} \) with respect to \( x \) is \( u'(x)=3x^{2} \) (using the power rule \( \frac{d}{dx}(x^{n})=nx^{n - 1} \)).

Step2: Apply the Fundamental Theorem and Chain Rule

First, evaluate \( g(u(x)) \). Substitute \( u(x)=x^{3} \) into \( g(t)=t^{4} \), we get \( g(u(x))=(x^{3})^{4} \). Using the power - of - a - power rule \( (a^{m})^{n}=a^{mn} \), \( (x^{3})^{4}=x^{12} \).

Then, by the chain rule, \( f'(x)=g(u(x))\cdot u'(x) \). Substitute \( g(u(x)) = x^{12} \) and \( u'(x)=3x^{2} \) into the formula:

\( f'(x)=x^{12}\cdot3x^{2} \)

Using the product rule for exponents \( a^{m}\cdot a^{n}=a^{m + n} \), we have \( x^{12}\cdot3x^{2}=3x^{12 + 2}=3x^{14} \).

Answer:

\( 3x^{14} \)