Question

let f be the function defined by $f(x) = \int_{4}^{\sqrt{3x}} \sin(5t) \\, dt$. find an expression for $f(x)$.

Explanation:

Step1: Recall the Fundamental Theorem of Calculus (Part 1) and Chain Rule

The Fundamental Theorem of Calculus, Part 1, states that if \( F(x)=\int_{a}^{g(x)}f(t)dt \), then \( F'(x) = f(g(x))\cdot g'(x) \), where \( a \) is a constant and \( g(x) \) is a differentiable function. Here, \( f(t)=\sin(5t) \), \( a = 4 \), and \( g(x)=\sqrt{3x}=(3x)^{\frac{1}{2}} \).

Step2: Find the derivative of \( g(x) \)

First, differentiate \( g(x)=(3x)^{\frac{1}{2}} \) with respect to \( x \). Using the power rule and chain rule, \( g'(x)=\frac{1}{2}(3x)^{-\frac{1}{2}}\cdot3=\frac{3}{2\sqrt{3x}} \). We can simplify \( \frac{3}{2\sqrt{3x}} \) by rationalizing or simplifying the square root: \( \frac{3}{2\sqrt{3x}}=\frac{\sqrt{3}}{2\sqrt{x}} \) (by multiplying numerator and denominator by \( \sqrt{3} \), since \( \frac{3}{\sqrt{3}}=\sqrt{3} \)).

Step3: Find \( f(g(x)) \)

Substitute \( t = g(x)=\sqrt{3x} \) into \( f(t)=\sin(5t) \). So, \( f(g(x))=\sin(5\sqrt{3x}) \).

Step4: Apply the Chain Rule to \( F(x) \)

Using the formula from the Fundamental Theorem of Calculus and Chain Rule, \( F'(x)=f(g(x))\cdot g'(x) \). Substitute \( f(g(x))=\sin(5\sqrt{3x}) \) and \( g'(x)=\frac{\sqrt{3}}{2\sqrt{x}} \) (or \( \frac{3}{2\sqrt{3x}} \)) into the formula.

So, \( F'(x)=\sin(5\sqrt{3x})\cdot\frac{3}{2\sqrt{3x}} \). We can also simplify \( \frac{3}{2\sqrt{3x}} \) as \( \frac{\sqrt{3}}{2\sqrt{x}} \), so \( F'(x)=\sin(5\sqrt{3x})\cdot\frac{\sqrt{3}}{2\sqrt{x}} \). Another way to write \( \sqrt{3x}= \sqrt{3}\sqrt{x} \), so \( \frac{3}{2\sqrt{3x}}=\frac{3}{2\sqrt{3}\sqrt{x}}=\frac{\sqrt{3}}{2\sqrt{x}} \) (since \( \frac{3}{\sqrt{3}}=\sqrt{3} \)).

Answer:

\( F'(x)=\frac{3\sin(5\sqrt{3x})}{2\sqrt{3x}} \) (or equivalently \( \frac{\sqrt{3}\sin(5\sqrt{3x})}{2\sqrt{x}} \))