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Chapter 6: Problem 17

$$ \text { Use integration by parts to find each integral. } $$ $$ \int(x-3)(x+4)^{5} d x $$

Short Answer

Expert verified
The integral is \( \frac{1}{6} x (x+4)^6 - \frac{1}{42} (x+4)^7 - \frac{1}{2} (x+4)^6 + C \).

Step by step solution

01

Expand the Integral

First, expand the integral by distributing \((x-3)\) inside the expression \((x+4)^{5}\). This gives us the expression: \[(x+4)(x+4)^{4} = x(x+4)^{5} - 3(x+4)^{5}\].Now, separate this into two individual integrals:\[\int x(x+4)^{5} \, dx - 3 \int (x+4)^5 \, dx\].
02

Integration by Parts on First Integral

Now apply integration by parts to the first integral:\(\int x(x+4)^{5} \, dx\).Set \(u = x\) and \(dv = (x+4)^{5} \, dx\).Differentiate and integrate to find \(du\) and \(v\):\(du = dx\) and \(v = \frac{1}{6}(x+4)^{6}\).Now substitute into the integration by parts formula:\[\int u \, dv = uv - \int v \, du\].This becomes:\[x \cdot \frac{1}{6}(x+4)^{6} - \int \frac{1}{6}(x+4)^{6} \, dx\].
03

Evaluate the Integral

Evaluate the second part of the integration by parts from Step 2:\[\int \frac{1}{6}(x+4)^{6} \, dx\]Integrate to find:\[\frac{1}{42}(x+4)^{7}\].
04

Solve Second Integral

Now solve the second integral, which was:\(-3 \int (x+4)^5 \, dx\).This becomes:\[ -3\cdot \frac{1}{6}(x+4)^{6} = -\frac{1}{2}(x+4)^{6}\] after simplifying.
05

Combine and Simplify

Now, combine all parts to get the solution to the original integral:\[\frac{1}{6}x(x+4)^{6} - \frac{1}{42}(x+4)^{7} - \frac{1}{2}(x+4)^{6}\].Simplify and combine like terms to show:\[\frac{1}{6} x (x+4)^6 - \frac{1}{42} (x+4)^7 - \frac{1}{2} (x+4)^6 + C\].Finally, collect coefficients to simplify the answer.

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Key Concepts

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

Definite Integrals
Definite integrals are essential concepts in calculus that involve evaluating integrals with specified upper and lower limits. Unlike indefinite integrals, which result in a general antiderivative plus a constant, definite integrals yield a specific numerical value representing the area under the curve between the two limits.
To solve definite integrals, we utilize the Fundamental Theorem of Calculus, which establishes the connection between differentiation and integration. Here's how you can approach solving definite integrals:
  • First, evaluate the indefinite integral of the function.
  • Next, substitute the upper limit into this antiderivative and calculate the result. Do the same for the lower limit.
  • Subtract the result of the lower limit from the result of the upper limit to find the value of the definite integral.
Understanding definite integrals allows you to calculate areas and manage problems involving accumulation, which are common in physics and engineering.
Polynomial Integration
Polynomial integration is a fundamental skill in calculus, as polynomials are simple yet powerful expressions that appear often in mathematical problems. The process involves finding the antiderivative, or integral, of polynomial expressions.
Each term of a polynomial can be integrated separately using the power rule for integration, which states that the integral of a term of the form \(ax^n\) is given by \(\frac{a}{n+1}x^{n+1}\)+ C, where \(C\) is the constant of integration.
Let's break it down further:
  • Identify each term in the polynomial and apply the power rule separately.
  • Add the results together for the complete integral.
  • If there are known boundary conditions or limits, apply them to find specific values.
By mastering polynomial integration, you gain the ability to solve a wide variety of calculus problems efficiently.
Calculus Techniques
Calculus techniques encompass a wide range of methods and tools to solve different mathematical problems, with integration by parts being one of these essential techniques. It is particularly useful when dealing with integrals that are products of functions.
The integration by parts formula, derived from the product rule of differentiation, is given by:\[\int u \, dv = uv - \int v \, du\]Here's how to apply it:
  • Select a part of the integrand as \(u\) such that its derivative, \(du\), simplifies the integral.
  • The other part becomes \(dv\), which you can integrate to find \(v\).
  • Substitute back into the formula to simplify and solve the integral.
Using techniques like substitution, partial fractions, and trigonometric integrals are equally important. Together, these calculus techniques enable you to tackle complex integrals, analyze functions, and solve real-world problems efficiently.

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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 \(y(t)\), where \(d\) and \(r\) are unknown constants, subject to the initial condition \(y(0)=0\) (zero initial value).

Evaluate each definite integral using integration by parts. (Leave answers in exact form.) $$ \int_{0}^{3} x e^{x} d x $$

Find in two different ways and check that your answers agree. \(\int x(x-2)^{5} d x\) a. Use integration by parts. b. Use the substitution \(u=x-2 \quad\) (so \(x\) is replaced by \(u+2\) ) and then multiply out the integrand.

Hospital patients are often given glucose (blood sugar) through a tube connected to a bottle suspended over their beds. Suppose that this "drip" supplies glucose at the rate of \(25 \mathrm{mg}\) per minute, and each minute \(10 \%\) of the accumulated glucose is consumed by the body. Then the amount \(y(t)\) of glucose (in excess of the normal level) in the body after \(t\) minutes satisfies $$y^{\prime}=25-0.1 y \quad \text { (Do you see why?) }$$ \(y(0)=0 \quad\) (zero excess glucose at \(t=0)\) Solve this differential equation and initial condition.

Suppose that you now have $$\$ 6000$$, you expect to save an additional $$\$ 3000$$ during each year, and all of this is deposited in a bank paying \(10 \%\) interest compounded continuously. Let \(y(t)\) be your bank balance (in thousands of dollars) \(t\) years from now. a. Write a differential equation that expresses the fact that your balance will grow by 3 (thousand dollars) and also by \(10 \%\) of itself. [Hint: See Example 7.] b. Write an initial condition to say that at time zero the balance is 6 (thousand dollars). c. Solve your differential equation and initial condition. d. Use your solution to find your bank balance \(t=25\) years from now.
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