Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{4} \frac{x-2}{\sqrt{x}} d x$$

Short answer

** The value of the definite integral is \(\frac{14}{3}\).

Step-by-step solution

Simplify the integrand

To begin, simplify the integrand by breaking the fraction into separate terms: $$\int_{1}^{4} \frac{x-2}{\sqrt{x}} d x = \int_{1}^{4} \left(\frac{x}{\sqrt{x}} - \frac{2}{\sqrt{x}}\right) d x$$ Now, rewrite the terms with exponents: $$\int_{1}^{4} \left(x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}\right) d x$$

Find the antiderivative

Next, find the antiderivative of the simplified integrand. Use the power rule for antiderivatives to integrate: $$F(x) = \int \left(x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}\right) d x = \frac{2}{3}x^{\frac{3}{2}} - 4x^{\frac{1}{2}} + C$$

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if F(x) is the antiderivative of the integrand, then: $$\int_{a}^{b} f(x) d x = F(b) - F(a)$$ Use the antiderivative found in step 2 and the limits of integration (1 to 4) to calculate the value of the definite integral: $$\int_{1}^{4} \left(x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}\right) d x = F(4) - F(1)$$

Evaluate the antiderivative at the limits of integration

Find the antiderivative's values at the limits of integration: $$F(4) = \frac{2}{3}(4)^{\frac{3}{2}} - 4(4)^{\frac{1}{2}} = \frac{16}{3} - 8$$ $$F(1) = \frac{2}{3}(1)^{\frac{3}{2}} - 4(1)^{\frac{1}{2}} = \frac{2}{3} - 4$$ Substitute these values into the Fundamental Theorem of Calculus expression: $$\int_{1}^{4} \left(x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}\right) d x = \left(\frac{16}{3}-8\right) - \left(\frac{2}{3}-4\right)$$

Calculate the value of the definite integral

Finally, simplify the expression to find the value of the definite integral: $$\int_{1}^{4} \left(x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}\right) d x = \left(\frac{16}{3} - 8\right) - \left(\frac{2}{3} - 4\right) = \frac{14}{3}$$ So, the value of the definite integral is \(\frac{14}{3}\).

Key concepts

Fundamental Theorem of Calculus

Understanding the Fundamental Theorem of Calculus (FTC) is essential for computing definite integrals accurately. In essence, the FTC bridges the concepts of differentiation and integration, two fundamental operations in calculus. The theorem has two parts, but the second part directly applies to evaluating definite integrals.

The FTC states that if a function is continuous on a closed interval \[a, b\] and F is its antiderivative, then the integral of the function over \[a, b\] is given by \[F(b) - F(a)\]. Put simply, you can find the total accumulation of a rate function (the integrand) over an interval by taking the difference of its antiderivative at the end points of the interval.

In the given exercise, applying the FTC allows us to easily find the definite integral of \(x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}\) from 1 to 4 by evaluating the antiderivative at the upper limit and subtracting the evaluation at the lower limit. This method is far more efficient than trying to sum an infinite number of infinitesimal pieces directly.

Antiderivative

An antiderivative can be thought of as the reverse operation to differentiation. For a given function f(x), an antiderivative is another function F(x) such that the derivative of F(x) is f(x), often denoted as \[F'(x) = f(x)\]. This is a core concept in integral calculus because the antiderivative is used to compute the area under the curve of a function, which is the integral.

When you find an antiderivative, you are essentially reversing the process of differentiation. However, there are infinitely many antiderivatives for any given function due to the constant of integration, denoted by 'C'. This constant represents the fact that when you take the derivative of a constant, the result is zero. Therefore, when integrating, you have to add this arbitrary constant to account for any possible shift along the y-axis.

In the problem given, the antiderivative \(F(x)\) of the function \(f(x) = x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}\) was found using the power rule for antiderivatives. This antiderivative \(F(x)\) is then used in the FTC to evaluate the definite integral.

Power Rule for Antiderivatives

The power rule for antiderivatives is a straightforward technique for integrating powers of x. It’s essentially the inverse of the power rule for derivatives. According to this rule, to find the antiderivative of \(x^n\), where n is any real number except -1, you increase the exponent by one and divide by the new exponent. In terms of an equation, for \(f(x) = x^n\), the antiderivative \(F(x)\) is given by \[F(x) = \frac{x^{n+1}}{n+1} + C\].

For the exercise at hand, the power rule was used to integrate \(x^{\frac{1}{2}}\) and \(x^{-\frac{1}{2}}\). The integrals of these terms were \(\frac{2}{3}x^{\frac{3}{2}}\) and \(\-4x^{\frac{1}{2}}\) respectively, following the power rule. Remember that integrating terms individually and then combining them is a valid and often used method known as the linearity of integrals. This technique is especially useful when dealing with polynomials or functions that can be broken down into simpler parts.