www.ck12.org Chapter 8. SE Integration Techniques - TI
C HAPTER
8 SE Integration Techniques - TI
Chapter Outline
8.1 INTEGRATION BY SUBSTITUTION
8.1. Integration by Substitution www.ck12.org
8.1 Integration by Substitution
This activity is intended to supplement Calculus, Chapter 7, Lesson 1.
In this activity, you will explore:
• Integration of standard forms
• Substitution methods of integration
Use this document to record your answers. Check your answers with theIntegratecommand.
Problem 1 –Introduction
1. Consider the integralR √
2x+3dx. Letu=2x+3. Evaluate the integral using substitution.
Use the table below to guide you.
T
ABLE8.1:
f(x) = √
2x+3
u= 2x+3
du= g(u) = Rg(u)du= R f(x)dx=
2. Try using substitution to integrateRsin(x)cos(x)dx. Letu=sin(x).
3. Now integrate the same integral, but letu=cos(x). How does the result compare to the one above?
4. sin(x)cos(x)dxcan be rewritten as12 sin(2x)using the Double Angle formula.
What is the result when you integrateR 12 sin(2x)using substitution?
Problem 2 –Common Feature
Find the result of the following integrals using substitution.
5.R x2+2x+3x+1 dx 6.Rsin(x)ecos(x)dx 7.R 4x2x+1dx
8. What do these integrals have in common that makes them suitable for the substitution method?
www.ck12.org Chapter 8. SE Integration Techniques - TI
Extension
Use trigonometric identities to rearrange the following integrals and then use the substitution method to integrate.
9.Rtan(x)dx 10. Rcos3(x)
8.2. Integration by Parts www.ck12.org
8.2 Integration by Parts
This activity is intended to supplement Calculus, Chapter 7, Lesson 2.
In this activity, you will explore:
• product rule of differentiation
• integration by parts
Use this document to record your answers.
Exercises
1. State the product rule for a function of the formu(x)∗v(x).
2. Apply the product rule to the function sin(x)∗ln(x).
3. Do you agree or disagree with the following statement? Explain.
Z d
dx(f(x))dx= d dx
Z
f(x)dx
= f(x)
4. What is the integral of the left side of the product rule?
Z
d
dx(u(x)·v(x)
dx=
5. What is the integral of the right side?
Z
u(x)·dv
dx+v(x)·du dx
dx=
6. Explain the relationship between the areas shown on the graph and the following equation:
v2
Z
v1
u·dv=u·v−
u2
Z
u1
v·du
www.ck12.org Chapter 8. SE Integration Techniques - TI
7. Use the method ofintegration by partsto compute the integral of ln(x).
Remember the formula for Integration by parts isRu·dv=u·v−Rv·du Z
ln(x)·1dx→u=ln(x)anddv=1dx
du= v=
Result=
Check by integration directly.(Home >F3:Calc >2:Integrate) or (Home >2nd7) Consider the function f(x) =sin(ln(x)).
u=sin(ln(x))→du=cos(ln(x))
x dx
dv=dx→v=x(+C)
Z
sin(ln(x))·1dx=x·sin(ln(x))−
Z
x·cos(ln(x))
x dx(+C)
=x·sin(ln(x))−
Z
cos(ln(x))dx(+C) 8. FindRcos(ln(x))dx.
u= du= dv= v=
Z
cos(ln(x))dx=
9. Substitute the result for cos(ln(x))into the result for sin(ln(x)).
8.2. Integration by Parts www.ck12.org
u= du= dv= v=
Z
sin(ln(x))dx=
10. Use integration by parts to solve the following. If you need to use integration by parts more than once, do so.
Check your result.
a.Rtan−1(x)dx b.Rx2·exdx c.Rx·tan−1(x)dx d.Rx·cos(2x+1)dx
11. (Extension 1)Does it matter in which orderu(x)andv(x)are selected for the method of integration by parts?
12. (Extension 2)Is there likely to be an integration rule based upon the quotient rule just as Integration by Parts was based upon the product rule?
www.ck12.org Chapter 8. SE Integration Techniques - TI
8.3 Charged Up
This activity is intended to supplement Calculus, Chapter 7, Lesson 7.
Part 1 –Separable Differential Equations Introduced
1. A capacitor, like one used for a camera flash, is charged up. When it discharges rapidly the rate of change of charge,q, with respect to time,t, is directly proportional to the charge. Write this as a differential equation.
The first step is to separate the variables, and then integrate and solve fory.
2. Findy(0), if dydx=sin(x)cos2(x)andy π2
=0. After integrating, use the initial conditiony=0 whenx=π2 to find the constant of integration. Then, substitutex=0 to findy(0).
Let’s return to the capacitor. Now that it is discharged, we need to get it charged up again.A9V battery is connected to a 100kΩresister,R, and 100µF capacitor,C.
The conservation of energy gives us the differential equation dqdt ·R=V−Cq → dqdt·R·C=V·C−q.
After substituting the given information and simplifying, we get the differential equation 10dqdt =0.9−q.
3. For the differential equation 10dqdt =0.9−q, separate the variables and integrate.
4. Apply the initial condition when time=0, the chargeq=0 and solve forq.
The syntax fordeSolveisdeSolve(y0= f(x,y),x,y)wherexis the independent andyis the dependent variable. The deSolvecommand can be found in theHOMEscreen by pressing2and selectingC:deSolve(.
5. On theHOMEscreen, typedeSolve(10q0=0.9−qandq(0) =0,t,q). Write down this answer and reconcile it with your previous solution.
6. In theHOMEscreen enterdeSolve(y0=y/x,x,y)to find the general solution of dydx=yx. Write the answer. Show your work to solve this differential equation by hand and apply the initial conditiony(1) =1 to find the particular solution.
Part 2 –Homework/Extension –Practice with deSolve and Exploring DEs
Find the general solution for the following separable differential equations. Write the solution in an acceptable format, (for example, useCinstead of@7). Show all the steps by hand if your teacher instructs you to do so.
1.y0=k·y 2.y0=xy 3.y0=2xy2
4.y0=3xy2
Open theGDBgraph labeled diffq1. Observe the family of solutions to the differential equation from Question 4, y0=3ty2. Many particular solutions can come from a general solution. When you are finished viewing the family of functions, go to theY=screen and delete the function.
8.3. Charged Up www.ck12.org 5. Not all differential equations are separable. Use deSolve to find the solution to the non-separable differential equationx·y0=3x2+2−y. What does this graph look like if the integration constant is 0? Explain. Open picture diffq2to view graph.
Find the particular solution for the following equations. Show your work. Solve fory. Explore other DEs on your own. Do you get any surprising results?
6.y0=x·y2andy(0) =1 7.y0=1+y2andy(0) =1 8.y0=7yandy(0) =ln(e)
www.ck12.org Chapter 9. SE Infinite Series - TI
C HAPTER
9 SE Infinite Series - TI
Chapter Outline
9.1 EXPLORINGGEOMETRICSEQUENCES