# In the problem below, use the Antiderivatives and Definite Integrals Theorem to evaluate the...

## Question:

In the problem below, use the Antiderivatives and Definite Integrals Theorem to evaluate the integral.

{eq}\displaystyle\; \int_{0}^{2} 4x^{3}\,dx, \quad\quad \int_{0}^{1} 4x^{3}\,dx, \quad\quad \int_{1}^{2} 4x^{3}\,dx {/eq}

## Definite integral:

The antiderivative of {eq}\displaystyle f(x) {/eq}, which is {eq}\displaystyle F(x) = \int f(x) \ dx {/eq}. The given integrand is a power function. To integrate, we'll use the integral power rule {eq}\displaystyle \int x^n \ dx = \dfrac{x^{n+1}}{n+1}+C. {/eq}

We use definite integral to compute many applications in mathematics. The fundamental theorem to evaluate the definite integral of {eq}f(x) {/eq} from {eq}a {/eq} to {eq}b {/eq} is {eq}\displaystyle \int_{a}^{b} f(x) \ dx= F(b)-F(a) {/eq}.

## Answer and Explanation:

We are given:

{eq}f(x) = 4x^3 {/eq}

First, we'll find out the antiderivative of {eq}f(x) = 4x^3 {/eq}, i.e {eq}\displaystyle F(x) =\int f(x) \ dx {/eq}

{eq}\displaystyle\Rightarrow F(x) =\int 4x^3 \ dx {/eq}

Take the constant out:

{eq}\displaystyle\Rightarrow F(x) =4 \int x^3 \ dx {/eq}

Apply integral power rule:

{eq}\displaystyle\Rightarrow F(x) =4 \dfrac{x^{3+1}}{3+1}+C {/eq}

{eq}\displaystyle\Rightarrow F(x) =4 \dfrac{x^{4}}{4}+C {/eq}

{eq}\displaystyle\Rightarrow F(x) =x^4+C {/eq}

**1.**

Compute {eq}\displaystyle\; \int_{0}^{2} 4x^{3}\,dx {/eq}

Use the fundamental theorem of definite integral {eq}\displaystyle\; \int_{0}^{2} 4x^{3}\,dx= F(2)-F(0) = (2^4-0^4)=16-0=16 {/eq}

Therefore, the solution is {eq}{\boxed{ \displaystyle\; \int_{0}^{2} 4x^{3}\,dx=16. }} {/eq}

**2.**

Compute {eq}\displaystyle\; \int_{0}^{1} 4x^{3}\,dx {/eq}

Use the fundamental theorem of definite integral {eq}\displaystyle\; \int_{0}^{1} 4x^{3}\,dx= F(1)-F(0) = (1^4-0^4)=1-0=1 {/eq}

Therefore, the solution is {eq}{\boxed{ \displaystyle\; \int_{0}^{1} 4x^{3}\,dx=1. }} {/eq}

**3.**

Compute {eq}\displaystyle\; \int_{1}^{2} 4x^{3}\,dx {/eq}

Use the fundamental theorem of definite integral {eq}\displaystyle\; \int_{1}^{2} 4x^{3}\,dx= F(2)-F(1) = (2^4-1^4)=16-1=15 {/eq}

Therefore, the solution is {eq}{\boxed{ \displaystyle\; \int_{1}^{2} 4x^{3}\,dx=15. }} {/eq}

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from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2