Problem 10 - Calculus

Question

Use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{1}^{4} \frac{s^{4}-8}{s^{2}} d s $$

Step-by-Step Solution

Answer

The value of the integral is 15.

Step 1: Simplify the Integrand

Begin by simplifying the integrand \( \frac{s^4 - 8}{s^2} \). This can be separated into two terms: \( \frac{s^4}{s^2} - \frac{8}{s^2} \) which simplifies to \( s^2 - 8s^{-2} \).

Step 2: Find the Antiderivative

The next step is to find the antiderivative of the simplified expression \( s^2 - 8s^{-2} \). The antiderivative of \( s^2 \) is \( \frac{1}{3}s^3 \), and the antiderivative of \( -8s^{-2} \) is \( 8s^{-1} \). Therefore, the antiderivative is \( \frac{1}{3}s^3 + 8s^{-1} + C \).

Step 3: Apply the Second Fundamental Theorem of Calculus

According to the Second Fundamental Theorem of Calculus, to evaluate a definite integral \( \int_{a}^{b} f(x) \, dx \), compute \( F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \). Here, evaluate \( \left[ \frac{1}{3}s^3 + 8s^{-1} \right]_{1}^{4} \).

Step 4: Evaluate the Antiderivative at the Upper Limit

Substitute \( s = 4 \) into the antiderivative \( \frac{1}{3}s^3 + 8s^{-1} \). This gives: \[ \left(\frac{1}{3}(4)^3 + 8(4)^{-1}\right) = \left(\frac{1}{3}(64) + \frac{8}{4} \right) = \left(\frac{64}{3} + 2 \right) = \frac{64}{3} + \frac{6}{3} = \frac{70}{3}. \]

Step 5: Evaluate the Antiderivative at the Lower Limit

Substitute \( s = 1 \) into the antiderivative \( \frac{1}{3}s^3 + 8s^{-1} \). This gives: \[ \left(\frac{1}{3}(1)^3 + 8(1)^{-1}\right) = \left(\frac{1}{3} + 8 \right) = \frac{1}{3} + \frac{24}{3} = \frac{25}{3}. \]

Step 6: Calculate the Definite Integral

Subtract the result from the lower limit from the upper limit: \[ \frac{70}{3} - \frac{25}{3} = \frac{45}{3} = 15. \] So, \( \int_{1}^{4} \frac{s^4 - 8}{s^2} \, ds = 15. \)

Key Concepts

Definite Integrals

A definite integral is a way to calculate the accumulation of quantities, like areas under a curve, between two points on the x-axis. It is written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. These limits define the interval over which the accumulation happens. In our example, we have \( a = 1 \) and \( b = 4 \).

• The lower limit \( a = 1 \) shows where the accumulation begins.
• The upper limit \( b = 4 \) shows where it ends.

To solve a definite integral using the Second Fundamental Theorem of Calculus, we need an antiderivative of the function \( f(x) \), which we'll evaluate at these limits and then calculate the difference.

Antiderivative

Finding the antiderivative is an essential step while evaluating definite integrals. An antiderivative of a function \( f(x) \) is another function \( F(x) \) such that the derivative \( \frac{d}{dx} F(x) = f(x) \). In simpler terms, this means reversing differentiation to uncover the original function.
For the integrand \( s^2 - 8s^{-2} \), we found that the antiderivative is \( \frac{1}{3}s^3 + 8s^{-1} + C \). Here's how:

• The antiderivative of \( s^2 \) is \( \frac{1}{3}s^3 \) because \( \frac{d}{ds} \frac{1}{3}s^3 = s^2 \).
• The antiderivative of \( -8s^{-2} \) is \( 8s^{-1} \) because \( \frac{d}{ds} 8s^{-1} = -8s^{-2} \).
• We add a constant \( C \), often omitted in definite integrals tasks, because any constant’s derivative is zero.

Simplifying Integrands

Simplifying the integrand can make finding antiderivatives much easier. The integrand is the function you are integrating, here written as \( \frac{s^4 - 8}{s^2} \). Before finding an antiderivative, it's necessary to simplify it.
This function can be broken down by dividing each term in the numerator by the denominator:

• \( \frac{s^4}{s^2} = s^2 \)
• \( \frac{-8}{s^2} = -8s^{-2} \)

Thus, the function simplifies to \( s^2 - 8s^{-2} \), which is much simpler to integrate. Breaking down complex problems into simpler parts can help make the integration process smoother and clearer.

Evaluating Limits

After finding the antiderivative, the next crucial step is evaluating it at the specific bounds—this results in determining the definite integral. The Second Fundamental Theorem of Calculus requires calculating \( F(b) - F(a) \), where \( F(x) \) is the antiderivative.
For this problem:

• Evaluate at the upper limit (\( s = 4 \)) gives us \( \frac{64}{3} + 2 = \frac{70}{3} \).
• Evaluate at the lower limit (\( s = 1 \)) results in \( \frac{1}{3} + 8 = \frac{25}{3} \).

To find the definite integral, subtract the lower limit's evaluation from the upper limit's: \( \frac{70}{3} - \frac{25}{3} = \frac{45}{3} = 15 \). This shows the area under the curve between \( s = 1 \) and \( s = 4 \) is 15. The process demonstrates the accumulation of the function's values over this interval.