Question

Evaluate each integral. $$\int_{25}^{225} \frac{d x}{\sqrt{x^{2}+25 x}}(\text { Hint: } \sqrt{x^{2}+25 x}=\sqrt{x} \sqrt{x+25})$$

Short answer

Question: Evaluate the definite integral $$\int_{25}^{225} \frac{dx}{\sqrt{x^{2}+25x}}.$$ Answer: The value of the definite integral is $$\int_{25}^{225} \frac{dx}{\sqrt{x^{2}+25x}} = 2\left(\frac{1}{5}\arctan 3 - \frac{1}{5}\arctan 1\right).$$

Step-by-step solution

Use the hint provided

Rewrite the given integral using the hint. $$\int_{25}^{225} \frac{dx}{\sqrt{x} \sqrt{x+25}}$$

Substitute \(u = \sqrt{x}\)

Let \(u = \sqrt{x}\), so \(x = u^2\). Taking the derivative of both sides with respect to \(x\), we get \(\frac{dx}{du} = 2u\). Now, substitute this back into the integral: $$\int_{5}^{15} \frac{2u\ du}{u(u^2 + 25)}$$ Note that we also have to change the limits of integration based on the substitution. When \(x = 25\), \(u = 5\), and when \(x = 225\), \(u = 15\).

Simplify and perform integration

Now, cancel \(u\) from the numerator and denominator: $$\int_{5}^{15} \frac{2\ du}{u^2 + 25}$$ To solve this integral, we use the formula: $$\int \frac{1}{x^2 + a^2}dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$ In our case, we have \(a = 5\) and \(x = u\). So our integral becomes: $$2\left(\frac{1}{5}\arctan\left(\frac{u}{5}\right)\right)$$ Evaluate the integral from \(5\) to \(15\): $$2\left(\frac{1}{5}\arctan\left(\frac{15}{5}\right) - \frac{1}{5}\arctan\left(\frac{5}{5}\right)\right)$$ $$= 2\left(\frac{1}{5}\arctan 3 - \frac{1}{5}\arctan 1\right)$$

Final answer

The value of the definite integral is: $$\int_{25}^{225} \frac{dx}{\sqrt{x^{2}+25x}} = 2\left(\frac{1}{5}\arctan 3 - \frac{1}{5}\arctan 1\right).$$

Key concepts

Integration Techniques

Understanding various integration techniques is crucial for solving complex calculus problems. The process of integration can be thought of as the reverse of differentiation. It involves finding a function, known as an antiderivative or integral, whose derivative is the given function. Integration techniques include methods such as substitution, integration by parts, partial fractions, and trigonometric integration.

Each method is used based on the form of the function being integrated. For instance, substitution is handy when an integral contains a composite function, whereas partial fractions are used when integrating rational functions with polynomial numerators and denominators. Mastering these techniques is essential as they often need to be combined to solve more challenging integrals.

U-Substitution

U-substitution is a powerful tool in integration, especially when dealing with integrals that involve composite functions. The idea is to simplify an integral by substituting the part of the function with a new variable, often denoted as 'u.' This simplification typically involves identifying a function and its derivative within the integral.

Once the substitution is made, the integral is often easier to solve. The change of variable transforms the original integral into one that is more manageable, after which the reverse substitution brings the solution back in terms of the original variable. It's also important to adjust the limits of integration when performing definite integrals, as shown in the given exercise with the change from 25 and 225 to 5 and 15 respectively, after the substitution.

Trigonometric Integration

Trigonometric integration refers to the integration of functions that involve trigonometric functions such as sine, cosine, tangent, and their reciprocals. This type of integration often involves using trigonometric identities to simplify the integral or transform it into a simpler form that can be integrated easily.

For functions involving \(x^2 + a^2\), the arctangent function often comes into play. The key is recognizing the standard forms and knowing their antiderivatives, such as \(\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C\). Therefore, an understanding of trigonometric identities and their derivatives is essential for trigonometric integration.

Arctangent Function

The arctangent function is the inverse of the tangent function and is a vital part of calculus, particularly in trigonometric integration. It appears whenever you have an integral of the form \(\frac{1}{x^2 + a^2}\) and serves as a function that 'undoes' what the tangent function does.

Mathematically, for a variable \(x\), the arctan of \(x\) gives you the angle whose tangent is \(x\). In integration, when a function's denominator is of the form \(x^2 + a^2\), the antiderivative involves \(\arctan\), as seen in the exercise. The arctangent function is particularly useful when solving definite integrals because it allows you to find precise values of the areas under curves represented by functions of this specific form.