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Chapter 5: Problem 29

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{2} 4 x^{3} d x$$

Short Answer

Expert verified
Answer: The value of the definite integral of the function \(4x^3\) from \(0\) to \(2\) is equal to \(16\).

Step by step solution

01

Find the antiderivative of the given function

In order to determine the antiderivative of the given function \(4x^3\), we need to apply the power rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\). So for our function, we have: $$\int 4x^3 dx = 4\int x^3 dx = 4\left(\frac{x^{3+1}}{3+1}\right) + C = x^4 + C.$$
02

Apply the Fundamental Theorem of Calculus

Now that we have found the antiderivative of the function, we can use the Fundamental Theorem of Calculus to evaluate the definite integral. This theorem states that the definite integral of a continuous function over an interval \([a, b]\) is given by the difference of the values of its antiderivative evaluated at the limits of integration: $$\int_{a}^{b} f(x) dx = F(b) - F(a),$$ where \(F(x)\) is the antiderivative of the function \(f(x)\).
03

Substitute the limits of integration and find the definite integral

In our case, the limits of integration are \(0\) and \(2\). Thus, substituting these limits into the Fundamental Theorem of Calculus equation, we get: $$\int_{0}^{2} 4x^3 dx = x^4|_{0}^{2} = (2^4)-(0^4) = 16 - 0 = 16.$$
04

State the final result

By applying the Fundamental Theorem of Calculus, we found that the definite integral of the given function \(4x^3\) from \(0\) to \(2\) is equal to \(16\). $$\int_{0}^{2} 4 x^{3} d x = 16.$$

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