Problem 13 - Calculus

Question

Use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{0}^{1}\left(2 x^{4}-3 x^{2}+5\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{22}{5} \).

Step 1: Identify the Function to Integrate

Our given integral is \( \int_{0}^{1} (2x^4 - 3x^2 + 5) \, dx \). This means that we need to evaluate the integral of the function \( f(x) = 2x^4 - 3x^2 + 5 \) from 0 to 1.

Step 2: Find the Antiderivative

According to the Second Fundamental Theorem of Calculus, if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \). We first need to find the antiderivative of \( f(x) = 2x^4 - 3x^2 + 5 \).

Step 3: Compute the Antiderivative

The antiderivative of \( f(x) = 2x^4 - 3x^2 + 5 \) is computed term by term:- The antiderivative of \( 2x^4 \) is \( \frac{2}{5}x^5 \).- The antiderivative of \( -3x^2 \) is \( -x^3 \).- The antiderivative of \( 5 \) is \( 5x \).Thus, the antiderivative \( F(x) \) is \( \frac{2}{5}x^5 - x^3 + 5x \).

Step 4: Evaluate the Definite Integral Using the Antiderivative

Now, use the antiderivative \( F(x) \) to evaluate the integral from 0 to 1:\[\int_{0}^{1} (2x^4 - 3x^2 + 5) \, dx = F(1) - F(0)\]

Step 5: Calculate \( F(1) \) and \( F(0) \)

Substitute \( x = 1 \) into \( F(x) \):\[ F(1) = \frac{2}{5}(1)^5 - (1)^3 + 5(1) = \frac{2}{5} - 1 + 5 \]Substitute \( x = 0 \) into \( F(x) \):\[ F(0) = \frac{2}{5}(0)^5 - (0)^3 + 5(0) = 0 \]

Step 6: Simplify the Result

Now compute \( F(1) - F(0) \):\[ F(1) - F(0) = \left( \frac{2}{5} - 1 + 5 \right) - 0 \]Simplify:\[ \frac{2}{5} - 1 + 5 = \frac{2}{5} + \frac{25}{5} - \frac{5}{5} = \frac{22}{5} \]Thus, the result of the definite integral is \( \frac{22}{5} \).

Key Concepts

Definite Integral

A definite integral is a powerful mathematical tool used to find the area under a curve with specific boundaries, often represented as \( \int_a^b f(x)\, dx \). Here, \( a \) and \( b \) denote the limits of the integral, establishing the interval over which the function \( f(x) \) is evaluated. This area provides insights into various applied scenarios such as total accumulated values and probability outcomes.
When working with definite integrals, you are essentially summing an infinite number of infinitesimally small quantities over the range from \( a \) to \( b \). It's a way of accumulating change across an interval.

• The function inside the integral, \( f(x) \), provides the height of the curve at each point \( x \).
• Integral bounds, \( a \) and \( b \), signal where to start and stop accumulating area.

If you've seen simple shapes, like rectangles or triangles, used to estimate areas, indefinite integral goes a step further, calculating it exactly using calculus tools.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function that 'undoes' differentiation. For a function \( f(x) \), an antiderivative \( F(x) \) satisfies the condition \( F'(x) = f(x) \).
Finding an antiderivative is akin to recognizing the original function from its derivative. Basic rules of integration—like the power rule—make this process systematic and mechanical for polynomials.

• To find an antiderivative of a polynomial like \( 2x^4 - 3x^2 + 5 \), integrate each term separately.
• The power rule for integration states that \( \int x^n \ dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

This rule allows us to compute antiderivatives like \( \frac{2}{5}x^5 - x^3 + 5x \) by adjusting exponents and coefficients according to the formula.

Evaluating Integrals

Evaluating integrals is often the culmination of calculus, where you apply the antiderivative formula over a specified range to find a precise value. When you evaluate a definite integral, the Second Fundamental Theorem of Calculus is your guiding principle.
This theorem elegantly states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_a^b f(x) \ dx = F(b) - F(a) \). With this key, you can unlock the total accumulation between the boundaries \( a \) and \( b \).

• Compute \( F(x) \) over the interval \([a, b]\) using the antiderivative found earlier.
• Calculate \( F(b) \), the value of the antiderivative evaluated at the upper limit.
• Calculate \( F(a) \), the value of the antiderivative at the lower limit, which often simplifies the problem if \( a = 0 \).
• Subtract \( F(a) \) from \( F(b) \) to find the total area under the curve.

As shown in the example, this process helped to identify that the total area is \( \frac{22}{5} \). Each step solidifies your mastery of evaluating integrals with precision and clarity.