Simple calculus integration help

In summary, the conversation discusses integrating with an upper bound of 2 and a lower bound of -3 for the function (4-y^2) - (y-2) dy. The books answer is 125/6, but the person gets 29.5/3 or 6. After some discussion and clarification, it is determined that there may have been a sign error in the integration process. It is also mentioned that reviewing previous math techniques may be helpful in understanding current material.
  • #1
Lavid2002
17
0

Homework Statement


Integrate with an upper bound of 2 and a lower bound of -3
(4-y^2) - (y-2) dy



Homework Equations





The Attempt at a Solution


The books answer is 125/6 I have tried twice and got 29.5/3 and 6 : /

This will be my second time taking calculus 2 and I always have a hard time getting back into the zone of differentiation and integration.

I first simplified the equation into upper bound of 2 lower bound of -3 6-y^2-y dy

Then I integrated it piece by piece.
6 turns into 6y,
-y^2 turns into -y^3 /3
-y turns into -y^2/2

Then I plug in 2 for all the y's to give me A
Then I plug in -3 for all the y's to give me B
Then I do A-B (Upper bound plugged in minus lower bound plugged in) and I get 29.5/3

What am I doing wrong?

Thanks
-Dave
 
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  • #2
Lavid2002 said:
I first simplified the equation ...
This is an integral, not an equation.

You are probably making a sign error somewhere, since your antiderivative is correct.
[tex]\left. 6y - \frac{y^2}{2} - \frac{y^3}{3}\right|_{-3}^2[/tex]
= (6(2) - 4/2 - 8/3) - (-18 - 9/2 + 27/3 )

If you are careful in evaluating this, you should get 125/6.
 
  • #3
Lavid2002 said:
Then I plug in 2 for all the y's to give me A
Then I plug in -3 for all the y's to give me B
Then I do A-B (Upper bound plugged in minus lower bound plugged in) and I get 29.5/3

What am I doing wrong?

Thanks
-Dave
You're probably making some algebraic errors. When I plug in 2 I get 22/3, and when I plug in -3 I get -27/2.
22/3 - (-27/2) = 125/6EDIT: Oops, beaten to it. ;)
 
  • #4
I got it now : /

This is what messes me up in this class. Not the calc II, the simple calc I and the algebra!
Very frustrating.

First homework assignment down.

Thanks bud
 
  • #5
Lavid2002 said:
This is what messes me up in this class. Not the calc II, the simple calc I and the algebra!
Very frustrating.
This is not unique to you. Unlike many other subjects, math builds on knowledge and skills from previous classes. If you find that you are having difficulties with techniques that were presented in earlier classes, set aside some time to spend reviewing those techniques and working a few problems. A small investment of time now will help you out a lot later on.
 

1. What is integration in calculus?

Integration in calculus is a mathematical process of finding the area under a curve. It is the inverse operation of differentiation, which is used to find the slope of a curve. Integration is an important tool in many fields of science, engineering, and economics.

2. How do you solve integration problems?

To solve an integration problem, you need to follow a series of steps. First, identify the function that you need to integrate and write it in integral form. Then, use integration rules and techniques, such as substitution, integration by parts, or trigonometric identities, to simplify the integral. Finally, evaluate the integral using the limits of integration and solve for the unknown variable.

3. What is the difference between indefinite and definite integration?

Indefinite integration involves finding a general solution to an integral without specific limits of integration. It results in an indefinite integral, also known as an antiderivative. On the other hand, definite integration involves finding the numerical value of an integral within specific limits of integration.

4. Can integration be used to find the volume of a three-dimensional shape?

Yes, integration can be used to find the volume of a three-dimensional shape in calculus. This process is known as triple integration, where you integrate a function of three variables over a three-dimensional region to find its volume. It is an important concept in multivariable calculus and is used in many applications, such as in engineering and physics.

5. What are the common applications of integration in science?

Integration has many applications in science, including physics, engineering, economics, and statistics. In physics, integration is used to calculate the work done by a force, the center of mass of an object, and the velocity and acceleration of an object. In engineering, integration is used in designing structures, calculating volumes and areas, and solving differential equations. In economics, integration is used to find the total profit or cost of a business. In statistics, integration is used to find the probability of events and to calculate expected values.

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