Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{2} \frac{2 s^{2}-4}{s^{3}} d s$$

Short answer

Question: Evaluate the integral \(\int_{1}^{2}\frac{2s^2-4}{s^3} ds\). Answer: The value of the definite integral is 6.

Step-by-step solution

Determine the Antiderivative

Begin by finding the antiderivative of the integrand, which is the function inside the integral: \(\frac{2s^2 - 4}{s^3}\). In order to do that, we will rewrite the integrand as a sum of separate fractions: $$\frac{2s^2 - 4}{s^3} = \frac{2s^2}{s^3} - \frac{4}{s^3} = 2 \cdot \frac{1}{s} - 4 \cdot \frac{1}{s^3} = 2s^{-1} - 4s^{-3}$$ Now we can find the antiderivative for each individual term: $$\int (2s^{-1} - 4s^{-3}) ds = 2\int s^{-1} ds - 4\int s^{-3} ds$$ Apply the power rule for integration: $$2[(-1+1)s^{-1+1}] - 4[(-3+1)s^{-3+1}] = 2 (s^{0}) - 4(-2s^{-2})= 2 - 8s^{-2}$$ The antiderivative is: $$F(s) = 2 - \frac{8}{s^2}$$

Apply the Fundamental Theorem of Calculus

Now that we have the antiderivative \(F(s)\), we can apply the Fundamental Theorem of Calculus to find the value of the integral. The theorem states that for a continuous function \(f(s)\) with antiderivative \(F(s)\): $$\int_{a}^{b} f(s) ds = F(b) - F(a)$$ In our case, \(a = 1\), \(b = 2\), and \(F(s) = 2 - \frac{8}{s^2}\). Plug these values into the formula: $$\int_{1}^{2} \frac{2s^2 - 4}{s^3} ds = F(2) - F(1)$$

Evaluate the Antiderivative at the Limits

Next, we have to evaluate the antiderivative, \(F(s)\), at the specified limits, \(1\) and \(2\). $$F(2) = 2 - \frac{8}{(2)^2} = 2 - \frac{8}{4} = 2 - 2$$ and $$F(1) = 2 - \frac{8}{(1)^2} = 2 - 8 = -6$$

Compute the Value of the Integral

Finally, subtract the values of the antiderivative at the limits: $$\int_{1}^{2} \frac{2s^2 - 4}{s^3} ds = F(2) - F(1) = (2 - 2) - (-6) = 0 + 6 = 6$$ Thus, the value of the definite integral is 6.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of calculus, linking the concept of differentiation with integration. It provides a practical way to evaluate definite integrals, which are used to find areas under curves, among other applications. According to the theorem, if you have a continuous function \( f(s) \) on a closed interval \([a, b]\), and \( F(s) \) is its antiderivative, then the definite integral of \( f(s) \) from \( a \) to \( b \) is equal to the difference \( F(b) - F(a) \). This essentially means:

• Find an antiderivative \( F(s) \) of the function that you’re integrating.
• Evaluate \( F(s) \) at the upper limit \( b \) and the lower limit \( a \).
• Subtract \( F(a) \) from \( F(b) \) to find the value of the integral.

In our exercise, this process was used to evaluate the integral of the function \( \int_{1}^{2} \frac{2s^2 - 4}{s^3} ds \), helping us find the total area under the curve between \( s = 1 \) and \( s = 2 \). The result was a value of \( 6 \), showing how the Fundamental Theorem of Calculus simplifies the computational task by shifting the focus from summing an infinite number of values to just computing two simple evaluations.

Antiderivative

An antiderivative, also known as an indefinite integral, is the reverse of taking a derivative. This means finding a function whose derivative gives you the original function you are integrating. In the process of solving definite integrals, finding the antiderivative is a crucial step.
A good way to understand this is through our problem, where we had to deal with the function \( \frac{2s^2 - 4}{s^3} \). To find its antiderivative, we first simplified the expression to \( 2s^{-1} - 4s^{-3} \). By integrating these terms separately, we calculated:

• \( \int 2s^{-1} ds = 2 \cdot \ln|s| \), but since we only looked at derivatives, the \( \ln|s| \) was simplified.
• \( \int (-4)s^{-3} ds = \frac{4}{2} s^{-2} = 2s^{-2} \).

The sum of these was our antiderivative \( F(s) = 2 - \frac{8}{s^2} \). After finding this, we used it to compute the actual values to find the integral using those boundaries \( a \) and \( b \) according to the Fundamental Theorem of Calculus.

Power Rule for Integration

The Power Rule for Integration is a basic yet powerful tool that simplifies the process of finding antiderivatives, especially when involving polynomials. Essentially, this rule states that the antiderivative of \( s^n \), where \( n eq -1 \), can be found using the formula:\[\int s^n \, ds = \frac{s^{n+1}}{n+1} + C\]where \( C \) is the constant of integration.
In our practical exercise, we applied this rule to each term of our integrand after rewriting \( \frac{2s^2 - 4}{s^3} \) into individual powers of \( s \). Specifically:

• For \( 2s^{-1} \), the antiderivative using a logarithmic concept is simplified until we find a suitable form.
• For \( -4s^{-3} \), using the power rule we found \( \frac{-4}{-3+1} s^{-3+1} = 2s^{-2} \).

This method simplifies each component, allowing us to find the complete antiderivative seamlessly. Remember to always increment the power by one and divide by the new power to use the Power Rule for Integration effectively.