Problem 98 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{-1}^{3}\left(2 x^{2}-1\right) d x $$

Step-by-Step Solution

Answer

The value of the definite integral is \( \frac{44}{3} \).

Step 1: Identify the integral expression

We need to evaluate the definite integral of the function \( f(x) = 2x^2 - 1 \) from \( x = -1 \) to \( x = 3 \). This is denoted as \( \int_{-1}^{3} (2x^2 - 1) \, dx \).

Step 2: Find the antiderivative

The antiderivative of \( 2x^2 \) is \( \frac{2}{3}x^3 \). The antiderivative of \( -1 \) is \( -x \). Therefore, the antiderivative of the entire expression \( 2x^2 - 1 \) is \( \frac{2}{3}x^3 - x \).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, evaluate the antiderivative at the upper limit and the lower limit, then subtract the two results. Calculate \( \left( \frac{2}{3}(3)^3 - 3 \right) - \left( \frac{2}{3}(-1)^3 - (-1) \right) \).

Step 4: Evaluate the expression at the upper limit

Substitute \( x = 3 \) into the antiderivative: \( \frac{2}{3}(3)^3 - 3 = \frac{2}{3}(27) - 3 = 18 - 3 = 15 \).

Step 5: Evaluate the expression at the lower limit

Substitute \( x = -1 \) into the antiderivative: \( \frac{2}{3}(-1)^3 + 1 = \frac{2}{3}(-1) + 1 = -\frac{2}{3} + 1 = \frac{1}{3} \).

Step 6: Subtract the results of the limits

Subtract the result from Step 5 from the result in Step 4: \( 15 - \frac{1}{3} = \frac{44}{3} \).

Step 7: Finalize the answer

The value of the definite integral \( \int_{-1}^{3} (2x^2 - 1) \, dx \) is \( \frac{44}{3} \).

Key Concepts

Antiderivative

When evaluating a definite integral, finding the antiderivative of the function you're working with is crucial. An antiderivative, also known as an indefinite integral, is a function whose derivative yields the original function. For example, if you have the function \( f(x) = 2x^2 - 1 \), its antiderivative \( F(x) \) is found by integrating each term separately:

- The antiderivative of \( 2x^2 \) is \( \frac{2}{3}x^3 \). This is because the general formula for integrating \( x^n \) is \( \frac{x^{n+1}}{n+1} \).
- The antiderivative of \(-1\) is \(-x\), since the derivative of \( -x \) is \(-1\).
- Combining these, the antiderivative for \( 2x^2 - 1 \) is \( \frac{2}{3}x^3 - x \).

This find-and-integrate process is known as "integration by term" and is essential for solving definite integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of differentiation with integration, showing that differentiation and integration are inverse processes. While it has a couple of parts, we often focus on the one relating to definite integrals. This part states that if \( F(x) \) is the antiderivative of \( f(x) \), then the definite integral from \( a \) to \( b \) is:
\\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
This means you first find the antiderivative \( F(x) \) of \( f(x) \), evaluate \( F(x) \) at the upper limit \( b \), then at the lower limit \( a \), and subtract the two results. In practical terms, for our exercise, we calculated \( \left( \frac{2}{3}(3)^3 - 3 \right) - \left( \frac{2}{3}(-1)^3 - (-1) \right) \) to find the answer. This deductive process allows you to transform calculus complexities into simpler arithmetic calculations.

Integral Evaluation

Evaluating a definite integral means calculating its exact value over a specific interval. This involves several steps that unfold as follows:

• Identify the integral to evaluate. In our exercise, it’s \( \int_{-1}^{3} (2x^2 - 1) \, dx \).
• Determine the antiderivative of the function within the integral, breaking it down term by term when necessary and simplifying it.
• Apply the Fundamental Theorem of Calculus. You substitute the upper and the lower limits into the antiderivative.
• Subtract the lower limit's result from the upper limit's result. For instance, substituting into our antiderivative resulted in: \( 15 - \frac{1}{3} = \frac{44}{3} \).

Understanding these steps allows you to tackle comprehensive calculus problems by organizing tasks into bite-sized operations. This systematic approach simplifies the evaluation, ensuring you arrive at the precise numerical result for an integral.