Question

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

Short answer

Question: Evaluate the definite integral of the function \(4x^{3}\) from \(0\) to \(2\). Answer: The definite integral of the function \(4x^{3}\) from \(0\) to \(2\) is \(16\).

Step-by-step solution

Find the antiderivative of the function

To find the antiderivative of the function \(4x^{3}\), we can apply the power rule of integration, which states that $$\int x^{n} d x = \frac{x^{n+1}}{n+1} + C$$ Where C is the constant of integration. In this case, \(n = 3\), so $$\int 4x^{3} d x = 4 \cdot \frac{x^{3+1}}{3+1} + C = x^{4} + C$$

Evaluate the definite integral using the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, we can now evaluate the definite integral as: $$\int_{0}^{2} 4 x^{3} d x = (x^{4} + C)\Big|_{0}^{2} = (x^{4} + C)\Big|_{0}^{2} = (2^{4} + C) - (0^{4} + C) = 2^{4} - 0^{4} = 16$$ So, the definite integral $$\int_{0}^{2} 4 x^{3} d x = 16$$

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a principle that connects differentiation with integration, two core operations in calculus. It essentially bridges the gap between antiderivatives and definite integrals. In simpler terms, it allows us to evaluate the area under a curve over a certain interval by taking the difference between the values of an antiderivative at the boundaries of the interval.

This powerful theorem has two main parts: The first part ensures that if a function is continuous over an interval, then its indefinite integral (antiderivative) is also continuous and differentiable on that interval. The second part, which is directly applied in the exercise, says that the definite integral of a function from point a to point b can be found by taking the difference of its antiderivative evaluated at these points.

Therefore, to calculate the definite integral \[\int_{a}^{b} f(x) dx\], you first find the antiderivative F(x) of the function f(x), and then compute \[F(b) - F(a)\]. This gives the net 'accumulated' value between a and b, which can represent a wide range of things such as area under the curve, total distance traveled, or the net change in quantity.

Power Rule of Integration

When dealing with integration, various rules can simplify the process of finding the antiderivative of a function. One of the most commonly used rules is the power rule of integration. This rule applies to functions where the variable x is raised to a power, which is a real number.

The power rule states that the integral of \[x^n\], where n is any real number other than -1, is given by \[\int x^n dx = \frac{x^{n+1}}{n+1} + C\], where C represents the constant of integration. The constant C is crucial because integration is the inverse operation of differentiation, and differentiating any constant gives 0, leaving us with an infinite number of antiderivatives.

In our exercise \[\int 4x^3 dx\], we apply the power rule by incrementing the exponent by 1 and then dividing by the new exponent. Multiplying by the constant 4, which comes from the original function, we get the antiderivative \[4 \cdot \frac{x^{4}}{4} + C = x^4 + C\], where C is the constant of integration. This rule is not only efficient but also easy to remember and apply, making it a fundamental tool for solving integration problems.

Antiderivative

In calculus, an antiderivative, also known as an indefinite integral, is a function that reverses what the derivative does. If F(x) is the antiderivative of f(x), then the derivative of F(x) is f(x). Symbolically, if \[F'(x) = f(x)\], then F(x) is an antiderivative of f(x). It's important to remember that there can be many antiderivatives for a given function because adding any constant to a function does not change its derivative.

For example, the antiderivative of \[2x\] could be \[x^2\], but it could also be \[x^2 + 3\], \[x^2 - 5\], or any other \[x^2 + C\] where C is a constant. The graph of each antiderivative will be a vertical translation of the graph of \[x^2\].

When solving for definite integrals as in the exercise, the constant of integration is not needed because it cancels out when taking the difference between the antiderivative at the upper and lower limits of integration, \[F(b) - F(a)\]. Nevertheless, understanding the concept of the antiderivative is crucial because it forms the foundation of many integral calculations, including the solution of our exercise.