Question

a. evaluate the definite integral.
$$\frac { d } { d x } \int _ { \frac { \pi } { 2 } } ^ { \sqrt { x } } \sin 13 t d t = \frac { d } { d x } \square$$
(simplify your answer. use integers or fractions for any num

Explanation:

Step1: Use the fundamental theorem of calculus and chain rule

Let \( F(t) \) be an antiderivative of \( \sin(13t) \), i.e., \( F^\prime(t)=\sin(13t) \). Then \( \int_{\frac{\pi}{2}}^{\sqrt{x}}\sin(13t)dt=F(\sqrt{x}) - F(\frac{\pi}{2}) \).

Step2: Differentiate the result from Step1

Differentiate \( F(\sqrt{x})-F(\frac{\pi}{2}) \) with respect to \( x \). By the chain rule, \( \frac{d}{dx}(F(\sqrt{x}))=F^\prime(\sqrt{x})\cdot\frac{1}{2\sqrt{x}} \), and \( \frac{d}{dx}(F(\frac{\pi}{2})) = 0 \) (since \( F(\frac{\pi}{2}) \) is a constant). Since \( F^\prime(t)=\sin(13t) \), when \( t = \sqrt{x} \), \( F^\prime(\sqrt{x})=\sin(13\sqrt{x}) \).

Answer:

\(\frac{\sin(13\sqrt{x})}{2\sqrt{x}}\)