Question

Verify the statement by showing that the derivative of the right side equals the integrand of the left side. $$ \int \frac{x^{2}-1}{x^{3 / 2}} d x=\frac{2\left(x^{2}+3\right)}{3 \sqrt{x}}+C $$

Short answer

The derivative of the right side of the given equation is \(\frac{x^{2}-1}{x^{3/2}}\), which is the same as the integrand of the left side, hence, the statement has been verified.

Step-by-step solution

Recognize The Right Side

The right side of the equation is \(\frac{2\left(x^{2}+3\right)}{3 \sqrt{x}}+C\). The task is to find its derivative.

Compute The Derivative

Using the quotient rule for derivative, the derivative \( d/dx[\frac{2\left(x^{2}+3 \right)}{3\sqrt{x}}] \) is computed as \(\frac{ \left(2 x + 0\right) \cdot 3 \sqrt{x} - \left[2\left(x^{2}+3\right)\right] \cdot \frac{3}{2 \sqrt{x}} }{ \left(3 \sqrt{x}\right)^{2} } \), which simplifies to \(\frac{x^{2}-1}{x^{3/2}}\).

Comparison

After computing the derivative, it's observed that the innovated derivative equals the integrand of the left side of the equation, which is \(\frac{x^{2}-1}{x^{3/2}}\). Hence, the given statement has been verified.

Key concepts

Quotient Rule Derivative

In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. If you have a function \( h(x) = \frac{f(x)}{g(x)} \) where both \( f(x) \) and \( g(x) \) are differentiable, the derivative \( h'(x) \) is given by \( h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).

For example, to find the derivative of \( \frac{2(x^2+3)}{3\sqrt{x}} \) using the quotient rule, we identify \( f(x) = 2(x^2+3) \) and \( g(x) = 3\sqrt{x} \) and their respective derivatives. Following the quotient rule, we get \( h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \) and after simplification, we obtain the original integrand \( \frac{x^2-1}{x^{3/2}} \). This methodical approach enables students to tackle complex differential problems in a step-by-step manner, ensuring clarity and a robust understanding of derivative operations.

Antiderivative

The antiderivative, also known as an indefinite integral, is the inverse operation to taking a derivative. If a function \( F(x) \) is an antiderivative of \( f(x) \) then, when \( F(x) \) is differentiated, the result will be \( f(x) \)—that is, \( F'(x) = f(x) \).

In the given exercise, the antiderivative of the function \( \frac{x^2-1}{x^{3/2}} \) is sought. To find this, a general formula or method of integration might be applied, such as integration by parts or substitution. The result \( \frac{2(x^2+3)}{3\sqrt{x}} + C \) includes the constant of integration \( C \) because antiderivatives are not unique: adding any constant to a function does not affect its derivative. Providing the antiderivative as part of the solution helps students see the relationship between differentiation and integration, reinforcing the fundamental theorem of calculus.

Integral Calculus

Integral calculus is the study of integrals and their properties. It is primarily concerned with the concept of accumulation and area under curves. For continuous functions, integrals are used to compute areas, volumes, and to determine the accumulated quantity from a rate of change.

In the context of our textbook exercise, integral calculus allows us to find the original function when given its rate of change. This process of finding the function based on its derivative (or differential) is known as finding the indefinite integral, or simply integration. The process involves several techniques, like the power rule, substitution method, or, in more complex scenarios, integration by parts.

The verification of the exercise demonstrates an application of integral calculus, where we use differentiation to show that a certain function is indeed the antiderivative of another. This interplay of integration and differentiation is one of the central themes in calculus, further enriching students' understanding of mathematical concepts and providing them with the tools to solve practical problems in science and engineering.