Question

In Exercises 9-12, evaluate \(\int_{C}\left(x^{2}+y^{2}\right) d s\) \(C: x\) -axis from \(x=0\) to \(x=3\)

Short answer

The solution to the integral \(\int_{C} (x^2+y^2) ds\) along \(C: x\)-axis from \(x=0\) to \(x=3\) is 9.

Step-by-step solution

Simplify the Integral

Start by taking advantage of the fact that the path \(C\) lies entirely on the x-axis. Therefore, everywhere along \(C\) it is true that \(y=0\). The integral can be simplified by replacing \(y^2\) with zero: \(\int_C (x^2 + 0) ds = \int_0^3 x^2 dx\).

Application of Fundamental Theorem of Calculus

The integral of \(x^2\) can be computed using the Fundamental Theorem of Calculus: the antiderivative of \(x^2\) is \(\frac{1}{3}x^3\). Thus, \(\int_0^3 x^2 dx = \left.\frac{1}{3}x^3\right|_0^3\).

Substituting the limits of the integral

Substitute the limits of the integral into the antiderivative. It states: \(\left.\frac{1}{3}x^3\right|_0^3 = \frac{1}{3}*3^3 - \frac{1}{3}*0^3\).

Evaluate the Expression

Finally, evaluate the expression to get the solution: \(\frac{1}{3}*27 - 0 = 9\).

Key concepts

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a profound connection between differentiation and integration, two principal concepts in calculus. It bridges the concept of derivative with that of the antiderivative or integral. There are two parts to this theorem, and in textbook exercises, it's most often the second part that students apply. This part states that if a function is continuous on the interval \[a, b\], and F is the antiderivative of f on this interval, then
\[\int_{a}^{b} f(x) dx = F(b) - F(a)\].
This can be read as the definite integral of a function f(x) over \[a, b\] is equal to the difference between the values of an antiderivative evaluated at b and a.

For example, in the given exercise, the student is asked to evaluate a line integral where C is along the x-axis, thereby simplifying the integral as we see that y=0. By invoking the fundamental theorem, the student can then find the definite integral of \(x^2\) from 0 to 3, effectively transitioning from the integral evaluation to subtracting the antiderivative's values at the boundaries of the interval.

Understanding Antiderivatives

An antiderivative, sometimes referred to as an indefinite integral, is a function F that reverses the process of differentiation of another function f. That is, if \(F'(x) = f(x)\), then F is an antiderivative of f. It's important to note that there are infinitely many antiderivatives for any given function, since adding a constant to F still gives a function that has the same derivative f. This constant is often represented as \(C\) and is the reason why indefinite integrals include a \(+ C\) term at the end.

In our exercise, the antiderivative of \(x^2\) is \(\frac{1}{3}x^3\), and we use this knowledge to evaluate the definite integral without the addition of any constant, as the limits of integration are specific values.

Integral Evaluation

Integral evaluation is the process of calculating the area under a curve, where the curve represents the function being integrated. When evaluating a definite integral, we're finding the net area between the function and the x-axis over a given interval, which in the context of the fundamental theorem, involves substituting the limits of integration into the antiderivative and calculating the difference.

In simpler terms, after you find an antiderivative of the function, you plug in the upper limit of the integration and subtract the value obtained by plugging in the lower limit. If you're working with a line integral, like in the provided exercise, you're particularly dealing with curves in space and the integral represents the combined effects along the path. Step-by-step, we move from understanding the path and simplifying the integral, to finding antiderivatives and their evaluation at specific points, concluding with calculating this difference to find the final value.