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Verify that F(x) is an antiderivative of the integrand f(x) and use Fundamental Theorem to...

Question:

Verify that {eq}F(x) {/eq} is an antiderivative of the integrand {eq}f(x) {/eq} and use Fundamental Theorem to evaluate the definite integral.

$$\int _1^4 \sqrt x \ dx, \ F(x) = \frac{2}{3}x^{3/2} $$

Fundamental Theorem of Calculus:

Any function f is continuous on given limits and F is antiderivative of f on given limits then integral can be written as

{eq}\displaystyle \int_p^q f(t) dt = F(q)-F(p). {/eq}

The primary consideration used to get the answer:

1. Radical rule: {eq}\displaystyle \sqrt{a}=a^{\frac{1}{2}}. {/eq}

2. The power rule of integration: {eq}\displaystyle \int x^adx=\frac{x^{a+1}}{a+1}, \quad a\ne -1. {/eq}

3. Take the constant out: {eq}\displaystyle \left(a\cdot f\right)'=a\cdot f'. {/eq}

4. The power rule of derivative: {eq}\displaystyle \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}. {/eq}

Answer and Explanation:

Solving for $$\displaystyle I = \int _1^4 \sqrt x \ dx $$

Use the radical rule.

$$\displaystyle = \int _1^4x^{\frac{1}{2}}dx $$

Apply the power rule.

$$\displaystyle = \left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]^4_1 $$

Simplify:

$$\displaystyle = \left[\frac{2}{3}x^{\frac{3}{2}}\right]^4_1 $$

Compute the boundaries.

$$\displaystyle = \left[\frac{2}{3}4^{\frac{3}{2}}-\frac{2}{3}1^{\frac{3}{2}}\right]^4_1 $$

Simplify:

$$\displaystyle = \frac{14}{3}. $$

Now, solving for $$\displaystyle F(x) = \frac{2}{3}x^{3/2} $$

Take derivative both sides.

$$\displaystyle \frac{d}{dx}F(x) = \frac{d}{dx}\left(\frac{2}{3}x^{\frac{3}{2}}\right) $$

Take the constant out.

$$\displaystyle F'(x) = \frac{2}{3}\frac{d}{dx}\left(x^{\frac{3}{2}}\right) $$

Apply power rule of derivative.

$$\displaystyle F'(x) = \frac{2}{3}\cdot \frac{3}{2}x^{\frac{3}{2}-1} $$

Simplify:

$$\displaystyle F'(x) = \sqrt{x} $$

Which is equal to {eq}f(x). {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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