Use the fundamental theorem of calculus and the chain rule to evaluate the following integral. (10
points)

d
dx Z x^2
a
√
t dt

To evaluate the given expression using the Fundamental Theorem of Calculus and the Chain Rule, let's start by breaking down the expression:

The expression is:

\[
\frac{d}{dx} \int_{a}^{x^2} \sqrt{t} \, dt
\]

### Step 1: Identify the structure
The Fundamental Theorem of Calculus, Part 1 states that if \( F(x) \) is an antiderivative of \( f(t) \), then:

\[
\frac{d}{dx} \int_{c}^{x} f(t) \, dt = f(x)
\]

However, in our problem, the upper limit of the integral is not just \( x \), but \( x^2 \). So we need to use the Chain Rule in conjunction with the Fundamental Theorem of Calculus.

### Step 2: Apply the Fundamental Theorem of Calculus
Define the function:

\[
G(x) = \int_{a}^{x^2} \sqrt{t} \, dt
\]

To find \( \frac{d}{dx} G(x) \), we first differentiate with respect to \( x^2 \) and then apply the chain rule:

\[
\frac{d}{dx} G(x) = \frac{d}{dx} \left( \int_{a}^{x^2} \sqrt{t} \, dt \right)
\]

Using the Fundamental Theorem of Calculus, we get:

\[
\frac{d}{dx^2} \left( \int_{a}^{x^2} \sqrt{t} \, dt \right) = \sqrt{x^2}
\]

### Step 3: Apply the Chain Rule
Now, applying the Chain Rule to account for the fact that the upper limit is \( x^2 \):

\[
\frac{d}{dx} G(x) = \sqrt{x^2} \cdot \frac{d}{dx}(x^2)
\]

Simplify this expression:

\[
\sqrt{x^2} = |x|
\]

and

\[
\frac{d}{dx}(x^2) = 2x
\]

Therefore:

\[
\frac{d}{dx} G(x) = |x| \cdot 2x = 2x \cdot |x|
\]

### Final Answer:

\[
\frac{d}{dx} \int_{a}^{x^2} \sqrt{t} \, dt = 2x |x|
\]

If \( x \geq 0 \), \( |x| = x \) so:

\[
2x \cdot x = 2x^2
\]

If \( x < 0 \), \( |x| = -x \) so:

\[
2x \cdot (-x) = -2x^2
\]

### Simplified expression:

\[
\frac{d}{dx} \int_{a}^{x^2} \sqrt{t} \, dt = 2x^2 \text{ if } x \geq 0, \text{ and } -2x^2 \text{ if } x < 0
\]

Would you like any more details or have any questions?

Here are some related questions you might find helpful:
1. How do you apply the Fundamental Theorem of Calculus to definite integrals?
2. What is the Chain Rule, and how is it used in differentiation?
3. How do you evaluate definite integrals with variable limits?
4. What are some practical examples of using the Fundamental Theorem of Calculus?
5. Can you explain how to differentiate composite functions?

**Tip:** When working with integrals that have variable limits, always remember to consider the effects of both the Fundamental Theorem of Calculus and the Chain Rule together!

Use the fundamental theorem of calculus and the chain rule to evaluate the following integral. (10 points)

d/dx ∫(from a to x^2) √t dt

Learn how to evaluate the integral d/dx ∫(from a to x^2) √t dt using the Fundamental Theorem of Calculus and the Chain Rule. This detailed solution walks through the key calculus concepts and formulas, ideal for college-level students studying Calculus I.

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