RaySix

Post

at November 1st 2023, 1:55:07 am.

Question:

Let f(x) be a function defined by

f(x) = \int_{0}^{x^2} \cos(t) \, dt

Determine the derivative f'(x).
Find the value of f'(2).
Let g(x) be a function defined as the definite integral of f(t) with respect to t from -3 to x, i.e., [ g(x) = \int_{-3}^{x} f(t) , dt ] Find g'(x) using the Fundamental Theorem of Calculus.

Solution:

The function f(x) represents the definite integral of \cos(t) from 0 to x^2. To find f'(x), we use the Fundamental Theorem of Calculus. According to the theorem, if a function F(x) is defined as an integral of another function f(t) with a variable limit, then the derivative of F(x) is equal to f(x).

Applying this theorem, we differentiate the integral with respect to x:

f'(x) = \cos(x^2) \cdot \frac{d}{dx} x^2

Since $\frac{d}{dx} x^2 = 2x$ , we have:

f'(x) = \cos(x^2) \cdot 2x

To find the value of f'(2), we substitute x = 2 into the expression we derived in part (1):

f'(2) = \cos(2^2) \cdot 2(2)

Simplifying:

f'(2) = \cos(4) \cdot 4

f'(2) = 4 \cos(4)

Therefore, the value of f'(2) is 4 \cos(4).

Given the function g(x) = \int_{-3}^{x} f(t) , dt, we need to find g'(x) using the Fundamental Theorem of Calculus. By the theorem, if G(x) is the antiderivative of g(x) such that G'(x) = g(x), then G(x) can be found by integrating f(t) with respect to t.

First, we determine the antiderivative of f(x):

F(x) = \int \cos(t) \, dt = \sin(t) + C

Next, using the Fundamental Theorem of Calculus, we substitute the limits of integration and evaluate g(x):

g(x) = F(x) \bigg|_{-3}^{x} = (\sin(x) + C) - (\sin(-3) + C)

g(x) = \sin(x) - \sin(-3)

Finally, to find g'(x), we differentiate g(x) using the chain rule:

g'(x) = \frac{d}{dx} (\sin(x) - \sin(-3))

g'(x) = \cos(x) - 0

Therefore, g'(x) = \cos(x).

This completes the solution to the problem.