Question

Use differentiation to match the antiderivative with the correct integral. [Integrals are labeled (a), (b), (c), and (d).] (a) \(\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x\) (b) \(\int \frac{\sqrt{x^{2}+16}}{x} d x\) (c) \(\int \sqrt{7+6 x-x^{2}} d x\) (d) \(\int \frac{x^{2}}{\sqrt{x^{2}-16}} d x\) $$ 4 \ln \left|\frac{\sqrt{x^{2}+16}-4}{x}\right|+\sqrt{x^{2}+16}+C $$

Short answer

The antiderivative \(4 \ln \left|\frac{\sqrt{x^{2}+16}-4}{x}\right|+\sqrt{x^{2}+16}+C\) corresponds to Integral (b) \(\int \frac{\sqrt{x^{2}+16}}{x} d x\).

Step-by-step solution

- Differentiate the Antiderivative

Differentiate the given antiderivative \(4 \ln \left|\frac{\sqrt{x^{2}+16}-4}{x}\right|+\sqrt{x^{2}+16}+C\) with respect to x to determine its corresponding integral.

- Compare with Available Integrals

The differentiation yields \(\frac{\sqrt{x^2+16}}{x}\). Compare this result with the available integrals.

- Match the Integral

The result of differentiation is only seen in Integral (b). Therefore, the provided antiderivative corresponds to this integral.

Key concepts

Differentiation

In mathematics, differentiation is a fundamental concept of calculus that deals with the determination of the rate at which a function changes at any given point. It is the process of finding the derivative of a function. The derivative represents the slope of the function's graph at a particular point and is a measure of how sensitive the function's output is to a change in its input value. In the context of the given exercise, differentiation is used as a tool to find which integral corresponds to the provided antiderivative.

To differentiate a function, we apply rules like the power rule, the product rule, the quotient rule, and the chain rule. Let's consider a simple example: to differentiate the function \( f(x) = x^2 \), we use the power rule which gives us \( f'(x) = 2x \). This derivative indicates that for each unit increase in \( x \), \( f(x) \) increases by \( 2x \) units.

Integral Calculus

While differentiation focuses on rates and slopes, integral calculus is all about accumulation and areas. The integral of a function offers information regarding the accumulation of quantities and helps in finding areas under curves, volumes, and much more. Integrals are often presented as the \'reverse operation\' of differentiation, and this concept is central to solving the given exercise.

There are two main types of integrals: indefinite and definite. An indefinite integral is a function that represents the antiderivative of the original function, often expressed with a '+C' to denote the constant of integration. For instance, the indefinite integral of \( f'(x) = 2x \) is \( f(x) = x^2 + C \). A definite integral, on the other hand, computes the accumulation between two specific bounds and is a number. In this exercise, we work with indefinite integrals to match them to their corresponding antiderivatives.

Antiderivative Matching

The process of antiderivative matching is like a puzzle where the goal is to connect each integral with its original function before differentiation — its antiderivative. This requires knowledge of differentiation to reverse engineer the given integrals. As seen in the exercise, we compute the antiderivative of the given function and then compare it to a list of integrals to find a match.

To improve the understanding of antiderivative matching, remember that the antiderivative is the function that was originally differentiated to get the derivative. One helpful strategy is to differentiate potential matches and see if the result matches our given antiderivative. By performing this test on each option, we can find the correct antiderivative-integral pair. In more intricate cases, visualization via a graph or understanding the nature of the function (whether it's an exponential, logarithmic, or trigonometric function) can be crucial in correctly matching antiderivatives with integrals.