Simple calculus integration help

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Homework Help Overview

The problem involves integrating the expression (4 - y²) - (y - 2) with specified bounds of -3 and 2. The original poster expresses difficulty in achieving the correct answer compared to the textbook solution.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to simplify the integral and integrate it piece by piece, but questions the accuracy of their results. Participants discuss potential algebraic errors and the evaluation of the antiderivative at the bounds.

Discussion Status

Participants are actively engaging in identifying possible errors in the original poster's calculations. Some guidance has been offered regarding the evaluation of the antiderivative, but no consensus has been reached on the exact nature of the mistakes.

Contextual Notes

The original poster indicates frustration with foundational algebra skills impacting their calculus performance, suggesting a need for review of earlier material.

Lavid2002
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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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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.
\left. 6y - \frac{y^2}{2} - \frac{y^3}{3}\right|_{-3}^2
= (6(2) - 4/2 - 8/3) - (-18 - 9/2 + 27/3 )

If you are careful in evaluating this, you should get 125/6.
 
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. ;)
 
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
 
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.
 

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