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43-48. For each definite integral: a. Evaluate it using the table of integrals on the inside back cover. (Leave answers in exact form.) b. Use a graphing calculator to verify your answer to part (a). $$ \int_{3}^{4} \frac{1}{x \sqrt{25-x^{2}}} d x $$

Short Answer

Expert verified
The integral evaluates to \( \left( \frac{1}{5} \right) \left( \sec^{-1} \frac{5}{4} - \sec^{-1} \frac{5}{3} \right) \).

Step by step solution

01

Identify the integral type

The given integral is \( \int_{3}^{4} \frac{1}{x \sqrt{25-x^{2}}} \, dx \). This resembles the form \( \int \frac{1}{a^2-x^2} \, dx \) found in the table of integrals.
02

Find the matching integral

Check the table of integrals to find a function that matches the integrand form \( \frac{1}{x \sqrt{a^2-x^2}} \). From the table, the integral \( \int \frac{1}{x \sqrt{a^2-x^2}} \, dx = \frac{1}{a} \sec^{-1} \frac{a}{x} + C \) looks like a match for values corresponding to \(a = 5\).
03

Apply the formula

Using the formula \( \int \frac{1}{x \sqrt{25-x^2}} \, dx = \frac{1}{5} \sec^{-1} \frac{5}{x} + C \), apply the limits 3 to 4. Calculate the definite integral: \[ \left[ \frac{1}{5} \sec^{-1} \frac{5}{x} \right]_{3}^{4} = \frac{1}{5} \sec^{-1} \frac{5}{4} - \frac{1}{5} \sec^{-1} \frac{5}{3} \].
04

Solve the expression

Compute \( \sec^{-1} \frac{5}{4} \) and \( \sec^{-1} \frac{5}{3} \) using the inverse secant function properties. Then evaluate: \[ \frac{1}{5} \left( \sec^{-1} \frac{5}{4} - \sec^{-1} \frac{5}{3} \right) \]. These can often be left in their inverse form due to table restrictions.
05

Verify using a calculator

Use a graphing calculator to approximate the values of \( \sec^{-1} \frac{5}{4} \) and \( \sec^{-1} \frac{5}{3} \) to verify the calculated definite integral value from Part (a). Ensure the calculator is set to the correct mode, either radians or degrees, based on interpretation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Evaluation
Definite integral evaluation can be seen as finding the area under a curve from one point to another on a graph. It involves integrating a function over a specific interval. In this exercise, you're working with the definite integral \( \int_{3}^{4} \frac{1}{x \sqrt{25-x^{2}}} \, dx \).

To evaluate this, you'll want to compare it to known formulae in the table of integrals. The form resembles \( \int \frac{1}{x \sqrt{a^2-x^2}} \, dx \), which evaluates to \( \frac{1}{a} \sec^{-1} \frac{a}{x} + C \) for \( a = 5 \). This technique simplifies finding antiderivatives significantly because you leverage pre-known results rather than computing from scratch.

Finally, applying limits during evaluation produces a numerical result representing the area between the curve, the x-axis, and the two limits—in our example, between x = 3 and x = 4.
Inverse Trigonometric Functions
Inverse trigonometric functions are crucial in calculus for handling integrals involving trigonometric relationships. In this case, the function \( \sec^{-1} \frac{a}{x} \) appears during integration. It represents the angle whose secant is \( \frac{a}{x} \).

When working with such functions, remember they adjust for values beyond the basic range and map these results back to angles.
  • Simplified: \( \sec^{-1} z \) gives you the angle \( \theta \) such that \( \sec(\theta) = z \).
  • Range: It primarily derives values from 0 to \( \pi \).
  • Application: Extensively used in converting complex fractions in calculus problems to recognizable angle measures.
While inverse trigonometric functions might feel abstract, they often connect directly to geometric interpretations valuable for real-world applications.
Graphing Calculators
Graphing calculators are tremendously useful tools for verifying calculus results and ensuring your manual calculations align with numerical estimations. In the exercise, after evaluating the integral using known formulas, a graphing calculator can confirm this by approximating \( \sec^{-1} \frac{5}{4} \) and \( \sec^{-1} \frac{5}{3} \).

Here’s how to effectively use one:
  • Set the correct mode by choosing between degrees and radians based on your requirements.
  • Graph the function and visually inspect the area under the curve for intuitive checks.
  • Use the numeric integration feature to compare results of manual calculations.
With these capabilities, graphing calculators stand as invaluable resources in both verification and exploration of mathematical problems.

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Most popular questions from this chapter

An annuity is a fund into which one makes equal payments at regular intervals. If the fund earns interest at rate \(r\) compounded continuously, and deposits are made continuously at the rate of \(d\) dollars per year (a "continuous annuity"), then the value \(y(t)\) of the fund after \(t\) years satisfies the differential equation \(y^{\prime}=d+r y\). (Do you see why?) Solve the differential equation above for the continuous annuity \(y(t)\) with deposit rate \(d=\$ 1000\) and continuous interest rate \(r=0.05\), subject to the initial condition \(y(0)=0 \quad\) (zero initial value).

Each equation follows from the integration by parts formula by replacing \(u\) by \(f(x)\) and \(v\) by a particular function. What is the function \(v\) ? $$ \int f(x) e^{x} d x=f(x) e^{x}-\int e^{x} f^{\prime}(x) d x $$

$$ \text { Use integration by parts to find each integral. } $$ $$ \int(x+a)^{n} \ln (x+a) d x \quad(n \neq-1) $$

Evaluate each definite integral using integration by parts. (Leave answers in exact form.) $$ \int_{0}^{4} z(z-4)^{6} d z $$

We omit the constant of integration when we integrate \(d v\) to get \(v\). Including the constant \(C\) in this step simply replaces \(v\) by \(v+C\), giving the formula $$ \int u d v=u(v+C)-\int(v+C) d u $$ Multiplying out the parentheses and expanding the last integral into two gives $$ \int u d v=u v+C u-\int v d u-C \int d u $$ Show that the second and fourth terms on the right cancel, giving the "old" integration by parts formula \(\int u d v=u v-\int v d u\). This shows that including the constant in the \(d v\) to \(v\) step gives the same formula. One constant of integration at the end is enough. Repeated Integration by Parts Sometimes an integral requires two or more integrations by parts. As an example, we apply integration by parts to the integral \(\int x^{2} e^{x} d x\). $$ \begin{array}{c} \int \underbrace{x^{2} e^{x} d x}_{u d v}=\underbrace{x^{2} e^{x}}_{u v}-\int \underbrace{e^{x} 2 x d x}_{w \atop d u}=x^{2} e^{x}-2 \int x e^{x} d x \\ {\left[\begin{array}{cc} u=x^{2} & d v & =e^{x} d x \\ d u=2 x d x & v=\int e^{x} d x=e^{x} \end{array}\right]} \end{array} $$ The new integral \(\int x e^{x} d x\) is solved by a second integration by parts. Continuing with the previous solution, we choose new \(u\) and \(d u\) :

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