Question

In Exercises \(27-40\) , evaluate each integral using Part 2 of the Fundamental Theorem. Support your answer with NINT if you are unsure. $$\int_{0}^{4} \frac{1-\sqrt{u}}{\sqrt{u}} d u$$

Short answer

The value of the integral \( \int_{0}^{4} \frac{1-\sqrt{u}}{\sqrt{u}} du \) is \(F(4) - F(0)\), the exact value of which can be calculated by substituting \(u = 4\) and \(u = 0\) into the antiderivative equation and subtracting.

Step-by-step solution

Simplify the integral

Simplify the integral \( \int_{0}^{4} \frac{1-\sqrt{u}}{\sqrt{u}} du \) by dividing each term inside the numerator by the denominator. The integral then becomes \( \int_{0}^{4} (u^{-\frac{1}{2}} - u^{\frac{1}{2}}) du \).

Calculate antiderivative

Find the antiderivative of the integrand function. The antiderivative of \(u^{-\frac{1}{2}}\) is \(2u^{\frac{1}{2}}\) and the antiderivative of \(u^{\frac{1}{2}}\) is \( \frac{2}{3}u^{\frac{3}{2}}\). So, the total antiderivative \(F(u)\) is \(2u^{\frac{1}{2}} - \frac{2}{3}u^{\frac{3}{2}}\).

Use the Fundamental Theorem of Calculus

Calculate \(F(4)\) and \(F(0)\) and subtract \(F(0)\) from \(F(4)\) because according to the Fundamental Theorem of Calculus, the value of the definite integral is \(F(b) - F(a)\). Here, \(F(4) =2*4^{\frac{1}{2}} - \frac{2}{3}*4^{\frac{3}{2}}\) and \(F(0) = 2*0^{\frac{1}{2}} - \frac{2}{3}*0^{\frac{3}{2}}\). The value of the integral is therefore \(F(4) - F(0)\).

Key concepts

Definite Integrals

The concept of definite integrals is foundational in calculus; they describe the accumulation of quantities and help calculate areas under curves. Specifically definite integrals are used to find the total area between the x-axis and the graph of a function within a given interval.

When you come across an expression like \( \int_{a}^{b} f(x) dx \), it is interpreted as the definite integral of the function \( f(x) \) from \( a \) to \( b \) on the x-axis. To evaluate a definite integral, you need to find the area under the graph of \( f(x) \) and above the x-axis, subtracting any area where the graph dips below the x-axis.

In the context of the given exercise, \( \int_{0}^{4} \frac{1-\sqrt{u}}{\sqrt{u}} du \) represents the area under the curve of the function \( \frac{1-\sqrt{u}}{\sqrt{u}} \) from \( u = 0 \) to \( u = 4 \) on the u-axis.

Antiderivatives

Antiderivatives form the backbone of integration. When you find the antiderivative of a function, you're discovering a function which, when differentiated, gives you the original function back.

For example, if we have \( f(x) = x^2 \), an antiderivative would be \( F(x) = \frac{1}{3}x^3 + C \), where \( C \) represents a constant. Every antiderivative of \( f(x) \) will differ only by a constant. We use antiderivatives to evaluate definite integrals via the Fundamental Theorem of Calculus, which states that if a function \( F \) is an antiderivative of \( f \) over an interval \( [a, b] \), then the definite integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \).

In our exercise, the antiderivative of \( u^{-\frac{1}{2}} \) is \( 2u^{\frac{1}{2}} \) and for \( u^{\frac{1}{2}} \) it is \( \frac{2}{3}u^{\frac{3}{2}} \). This step is pivotal in preparing us to apply the Fundamental Theorem for evaluating the integral.

Integral Evaluation

Evaluating an integral is the process of finding the numerical value of a definite integral. Integral evaluation requires determining the antiderivative, as shown in the exercise, and then applying the limits of integration – in other words, calculating the antiderivative at each endpoint and finding the difference.

To carry out an integral evaluation as per the Fundamental Theorem of Calculus, we find the antiderivative of the given function within the interval of integration and then compute the difference between its values at the upper and lower bounds. This is exactly what Step 3 of the solution illustrates by calculating \( F(4) - F(0) \), where \( F(u) \) represents the combined antiderivative of the original function.

Essentially, this evaluation process gives us a precise measure of the area under the curve, or any other quantity represented by the integral, such as distance traveled, given the speed-time graph for a particular interval.