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An antiderivative of a function is another function whose derivative is the original function you started with. Essentially, it's like "reverse-differentiating." For any function you wish to integrate, finding the antiderivative is the initial step.

In our exercise, we handle the function \(6e^{s/2}\). To find its antiderivative, recall the rule for integrating exponential functions, \(e^{ks}\), which states that the integral is \(\frac{1}{k}e^{ks}\). Here, the constant \(k\) is \(\frac{1}{2}\), making the antiderivative of \(e^{s/2}\) equal to \(2e^{s/2}\). Therefore, for \(6e^{s/2}\), multiplying by 6 gives \(12e^{s/2}\).

• Antiderivative of \(e^{ks}\) is \(\frac{1}{k}e^{ks}\).
• Adjust by constants like 6 to find the full antiderivative, which results in multiplying the antiderivative by 6.

Finding antiderivatives requires practice, especially determining the correct constants that affect the entire function.

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, providing a way to evaluate definite integrals via antiderivatives. This theorem simplifies the process significantly.

According to this theorem, if you have a continuous function \(f\) and its antiderivative \(F\), the definite integral of \(f\) from \(a\) to \(b\) can be evaluated by \(F(b) - F(a)\). This eliminates the constant term \(C\) you normally add when finding an indefinite integral because it cancels out when subtracting the antiderivative values.

• First compute \(F(b) = 12e^{b/2}\).
• Then compute \(F(a) = 12e^{a/2}\).
• Finally, subtract: \(F(b) - F(a)\) to get your definite integral.

In our problem, \(F(2) = 12e\) and \(F(-2) = 12/e\), giving the result as \(12(e - 1/e)\). Powerful, simple, and crucial for calculus studies.

Exponential functions have a constant base raised to a variable exponent, taking the form \(ae^{bx}\). Here, \(a\) controls the function's amplification, whereas \(b\) affects how quickly the function changes.

Typically, integration involving exponential functions follows a straightforward pattern given by their easy-to-identify form. In our example, \(6e^{s/2}\), the base \(e\) stays consistent while the variable \(s\) in the exponent alters solutions based on what's multiplied (\(1/2\) in this case).

• Recognize forms: \(ae^{bx}\) leads to solutions of \(\frac{a}{b}e^{bx}\).
• Adjust for multipliers, as we saw with the leading 6 becoming 12 due to \(\frac{1}{1/2}\) turning into 2.

Understanding how exponential functions behave both in and out of integration helps tackle complex calculus problems, turning them into manageable tasks.