Question

Find the area of the region bounded by the curve \(f(x)=\left(16+x^{2}\right)^{-3 / 2}\) and the \(x\) -axis on the interval [0,3]

Short answer

The area of the region is \(A = \frac{1}{2} - \frac{1}{4\sqrt{5}}\).

Step-by-step solution

STEP 1: Find the definite integral

We need to find the definite integral of the function \((16+x^2)^{-\frac{3}{2}} \) on the interval [0,3]. This is written as: \(\int_{0}^{3} (16+x^2)^{-\frac{3}{2}}dx\) We'll use substitution to evaluate this integral. Let \(u = 16 + x^2\). Then, \(\frac{du}{dx} = 2x\). Thus, \(x\, dx = \frac{1}{2} du\). Let's substitute these into our integral: \(\frac{1}{2} \int (u)^{-\frac{3}{2}} du\)

STEP 2: Integrated function

Now, let's find the integral of our new function with respect to u: \(\frac{1}{2} \int (u)^{-\frac{3}{2}} du = (-\frac{1}{4} u^{-\frac{1}{2}} + C)\) Now, substitute back for u: \(-\frac{1}{4}(16+x^2)^{-\frac{1}{2}} + C\)

STEP 3: Evaluate the definite integral

Now, let's plug the limits of our integral, 0 and 3, back into our antiderivative and find the difference: \(-\frac{1}{4}(16+3^2)^{-\frac{1}{2}} - (-\frac{1}{4}(16+0^2)^{-\frac{1}{2}}) \) Solve this expression: \(-\frac{1}{4}(25)^{-\frac{1}{2}} - (-\frac{1}{4}(16)^{-\frac{1}{2}})\) \(=\frac{1}{4}(4^{-\frac{1}{2}}) - \frac{1}{4}(5^{-\frac{1}{2}})\) \(=\frac{1}{4}(2^{-1}) - \frac{1}{4}(5^{-\frac{1}{2}})\) \(=\frac{1}{2} - \frac{1}{4\sqrt{5}}\)

STEP 4: Final answer

The area of the region bounded by the curve \(f(x)=(16+x^2)^{-\frac{3}{2}} \) and the x-axis on the interval [0, 3] is: \(A = \frac{1}{2} - \frac{1}{4\sqrt{5}}\)

Key concepts

Integration by Substitution

The method of integration by substitution is a powerful tool in calculus, particularly when dealing with integrals that are not in a standard form. It is often referred to as the reverse chain rule because it involves undoing the chain rule for derivatives.

Here’s a simple way to understand it: Suppose you have a complicated function inside an integral that you’d rather not deal with directly. By substituting this problematic piece with a simpler symbol—typically, the letter 'u'—you can transform the integral into one that’s easier to handle.

In the case of our exercise, the substitution chosen is u = 16 + x^2, where du = 2x dx. This choice is strategic because it simplifies the integral into one with a power of ‘u’ that is more straightforward to integrate. After integrating with respect to ‘u’, we must remember to substitute back in terms of ‘x’ to complete the solution. This powerful technique often simplifies the integral and is a cornerstone strategy in finding indefinite and definite integrals alike.

Indefinite Integral

An indefinite integral represents a family of functions with an added constant of integration ‘C’. When integrating a function, we're essentially finding the antiderivative that can generate the original function through differentiation.

Consider it as the reverse process of taking a derivative: just as derivatives represent rates of change, indefinite integrals offer a collection of functions that describe accumulated quantities where the rate of change is known.

In our exercise, we computed an indefinite integral after the substitution process. Note that we integrate without limits since we're identifying the general form of the antiderivative before applying it to the specific scenario required – in this case, the area under a curve within an interval.

Area Under a Curve

Calculating the area under a curve between a curve and the x-axis over a given interval is one of the most common applications of definite integral calculus. This concept is not just a mathematical abstraction; it has practical implications in various fields like physics and economics.

The definite integral provides us with this area, and it can be visualized as the sum of infinitely small rectangles extending from the curve down to the x-axis. By adding up these 'slices', you’re able to calculate the total area covered.

In our textbook problem, the final step after integration involved plugging the interval limits into the antiderivative. By taking the difference of the values at these limits, we determine the precise area between the function f(x) = (16 + x^2)^{-3/2} and the x-axis, from x = 0 to x = 3. The area is not just a number but a solution that connects the antiderivative to a geometric interpretation, which is vital in many scientific and engineering contexts.