Question

v\(\int_{0}^{3} \frac{x}{\sqrt{9-x^{2}}} d x\)

Short answer

The integral evaluates to \(-3\).

Step-by-step solution

Identify Substitution

The integral \( \int \frac{x}{\sqrt{9-x^2}} \, dx \) makes use of a trigonometric substitution. Notice the expression \( \sqrt{9-x^2} \), which suggests a substitution involving trigonometric functions. Here, we can use the substitution \( x = 3 \sin \theta \), since \( \sqrt{9-x^2} = 3\cos \theta \).

Differentiate Substitution

Differentiate the substitution \( x = 3 \sin \theta \) with respect to \( \theta \): \( dx = 3 \cos \theta \, d\theta \).

Substitute and Simplify

Substitute \( x = 3 \sin \theta \) and \( dx = 3 \cos \theta \, d\theta \) into the integral. It becomes:\[\int \frac{3 \sin \theta}{3 \cos \theta} \cdot 3 \cos \theta \, d\theta = 3 \int \sin \theta \, d\theta.\]The \( 3 \cos \theta \) terms cancel out.

Integrate with Respect to \( \theta \)

The integral \( 3 \int \sin \theta \, d\theta \) can be solved as:\[3 \int \sin \theta \, d\theta = -3 \cos \theta + C.\]

Convert Back to x

Convert \( \theta \) back to \( x \). Since \( x = 3 \sin \theta \), we have \( \sin \theta = \frac{x}{3} \), thus \( \theta = \arcsin \left( \frac{x}{3} \right) \) and using \( \sin^2 \theta + \cos^2 \theta = 1 \) it follows \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left( \frac{x}{3} \right)^2} = \frac{\sqrt{9-x^2}}{3}\). sub back:\[-3 \cos \theta = - \frac{\sqrt{9-x^2}}{1}.\]

Apply Limits

Now apply the limits of the definite integral, changing them according to \( x = 3 \sin \theta \). Therefore, when \( x = 0 \), \( \theta = 0 \) and when \( x = 3 \), \( \theta = \frac{\pi}{2} \). Substitute these limits:\[\left[ -3 \cos \theta \right]_{0}^{\frac{\pi}{2}} = -(3 \cdot 0) \, - \, (3 \cdot 1) = -3 \]

Final Result

Thus the result of the definite integral \( \int_0^3 \frac{x}{\sqrt{9-x^2}} \, dx \) is \(-3\).

Key concepts

Definite Integrals

Definite integrals are a critical concept in calculus that measure the overall accumulation of values, such as areas under a curve, over an interval. The definite integral of a function from point \( a \) to point \( b \) is represented as \( \int_a^b f(x) \, dx \). Here, \( f(x) \) is the function being integrated, and \( a \) and \( b \) specify the interval over which the integration occurs.

In solving definite integrals, one should:

• Identify the function to be integrated.
• Perform the integration to find the antiderivative.
• Evaluate the antiderivative at the upper and lower limits, and subtract these values to find the result.

In our example, we evaluated \( \int_0^3 \frac{x}{\sqrt{9-x^2}} \, dx \) using trigonometric substitution. The definite integral result was found to be \(-3\), showing the value accumulated by the function over that interval.

Trigonometric Identities

Trigonometric identities play an essential role in calculus, particularly in the substitution method for integration. Certain integrals can be simplified using these identities by substituting variables, allowing for more straightforward integration. Common identities include:

• \( \sin^2 \theta + \cos^2 \theta = 1 \).
• \( \tan^2 \theta + 1 = \sec^2 \theta \).

For the given problem, the substitution \( x = 3 \sin \theta \) led to simplifications involving the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). By substituting \( x \) with \( 3 \sin \theta \), the expression \( \sqrt{9-x^2} \) transformed into \( 3 \cos \theta \).

These identities are crucial for transforming complex integrals into simpler ones, aiding in easier computation of the definite integral.

Calculus Integration Techniques

Various integration techniques in calculus can simplify finding antiderivatives and calculating areas. One such technique is **substitution**, which involves changing the variable of integration to simplify the process. A special case of substitution is trigonometric substitution, useful for specific integrals involving square roots of quadratic expressions.

In our example, we performed trigonometric substitution, where \( x = 3 \sin \theta \) was used. This transformed the integral into a form easier to handle: \( \int \sin \theta \, d\theta \).

Other techniques include:

• **Integration by Parts**: Useful for products of functions, following the model \( \int u \, dv = uv - \int v \, du \).
• **Partial Fraction Decomposition**: Breaks a complex rational function into simpler fractions for easier integration.

Mastering these techniques enhances problem-solving skills in calculus and makes complex integrations manageable.