Question

In Exercises \(23-28,\) use areas to evaluate the integral. $$\int_{a}^{b} 2 s d s, \quad 0 < a < b$$

Short answer

The definite integral of \(2s\) from \(a\) to \(b\) is \(b^2 - a^2\).

Step-by-step solution

Find the integral

To find the integral of \(2s\), apply the power rule for integration which states that the integral of \(s^n\) is \(\frac{s^{n+1}}{n+1}\), where \(n \neq -1\). In this case, the integral of \(s\) is \(\frac{s^2}{2}\) and with an additional factor of 2, the integral of \(2s\) is \(2\cdot \frac{s^2}{2} = s^2\).

Apply the limits of integration

To find the definite integral, subtract the value of the indefinite integral at the lower limit from its value at the upper limit. Using the theorem, the definite integral of \(2s\) from \(a\) to \(b\) is \(b^2 - a^2\).

Key concepts

Power Rule for Integration

Understanding the power rule for integration is essential for solving integrals involving polynomial expressions. When encountering an integral such as \(\int s^n ds\), where \(n\) is a constant that's not equal to \( -1 \), this rule simplifies the process by increasing the exponent by one and dividing by the new exponent. Mathematically, the rule is expressed as \(\int s^n ds = \frac{s^{n+1}}{n+1} + C\), where \(C\) represents the constant of integration for indefinite integrals.

Let's apply this to our exercise where the integrand is \(2s\). We notice that \(2s\) is equivalent to \(2\cdot s^1\), and hence, by applying the power rule, we get \(\int 2s\,ds = 2\cdot\int s^1\,ds = 2\cdot\frac{s^{1+1}}{1+1}\). This simplifies to \(s^2\), which is the antiderivative needed to evaluate the definite integral.

Limits of Integration

The limits of integration are vital in calculating the value of a definite integral. They specify the range over which the function is integrated and often represent the boundaries of the area under the curve. The lower limit of integration is denoted as \(a\) and the upper limit as \(b\). For the definite integral \(\int_{a}^{b} f(s)\,ds\), we evaluate the antiderivative of \(f(s)\) at \(b\) and subtract the antiderivative evaluated at \(a\).

In our example, we are finding \(\int_{a}^{b} 2s\,ds\) with the antiderivative \(s^2\). Plugging in the upper and lower limits, we have \(b^2\) for the upper limit and \(a^2\) for the lower limit. The definite integral is \(b^2 - a^2\), representing the difference in the area from \(a\) to \(b\) under the curve of the function \(2s\).

Area Under a Curve

The concept of the area under a curve is fundamental in integral calculus. A definite integral like \(\int_{a}^{b} f(s)\,ds\) not only measures the accumulation of quantities but also provides the exact area under the curve \(f(s)\) between the vertical lines \(s = a\) and \(s = b\), above the \(s\)-axis.

For the task at hand, the function \(2s\) forms a straight line, which means the area under this curve between \(s = a\) and \(s = b\) forms a trapezoid. The definite integral computes this area exactly, which is particularly straightforward for linear functions like this one. However, this concept extends to more complex functions as well, where the area might not have a simple geometric shape. Integrals can compute the area under curves that are much more intricate, and this is one of the many powerful applications of integral calculus.