# 4 Integral Questions

• Dec 13th 2009, 05:36 PM
araym1
4 Integral Questions
few integration questions that are bugging me i have made attempt on all of them some of them are not even worth mentioning how i have tried it because it got me no where near the answer.

1. i am not even sure how to begin this one.

$
=\int_3^7 \sqrt(16t^2 + 8t + 1)dt
$

attempt:

$=\int_3^7 (16t^2 + 8t + 1)^1/2 dt$
$=\int_3^7 \frac {(16t^2 + 8t + 1)^3/2}{3/2} dt$
$=\int_3^7 \frac {2}{3} (16t^2 + 8t + 1)^3/2 dt$

or other path i have tried as well... but wasn't sure how to exactly fully proceed

$u = 16t^2 + 8t + 1$
$du = 32t + 8 dt$

so i have tried this then...

$\frac {1}{32t} + \frac {1}{8} du = dt$

2. this one i am not sure if i copied wrong or something but when i attempt it i get the answer of 1

question:
$
=\int_0^1 \frac{dx}{\sqrt(3 - 2x)}
$

attempt:

$
u = (3-2x)
$

$
du = -2dx
$

$
-\frac{1}{2}du = dx
$

$
=-\frac{1}{2}\int_0^1 u^-1/2 du = -\frac{1}{2}\int_0^1 \frac {u^1/2}{1/2} = -\frac{1}{2} * \frac{2}{1}(3-2x)^1/2
$

$
= -1 \sqrt (3-2x)
$

3. this one gives me some problems as well. question says
find the equation that passes thru (1,0) is tangent to the line $6x + y = 6$ and has $y" = \frac {12}{x^3}$

answer: $\frac {6 - 6x}{x}$

4. last question says
find first quadrant area of $x + y + y^2 = 2$

answer: $\frac {7}{6}$

attempt:
first rewrite the equation to be

$x = y + y^2 - 2$

got the limits using table of values and got result of (2,1)

$A = \int_1^2 x dy$

$= \int_1^2 (y + y^2 - 2)dy$

$= \int_1^2 (\frac{y^2}{2} + \frac {y^3}{3} - 2y)$

and then i am not sure if i am on the right path because it does not work out to correct answer.
• Dec 13th 2009, 05:49 PM
TKHunny
On #1, you seem to have forgotten the Chain Rule. Try completing the square under the radical and give some thought to a trigonometric substitution.
• Dec 13th 2009, 05:51 PM
TKHunny
On #2, you seem to have forgotten to change your limits when you changed variables.
• Dec 13th 2009, 05:53 PM
VonNemo19
[quote=araym1;424449]few integration questions that are bugging me i have made attempt on all of them some of them are not even worth mentioning how i have tried it because it got me no where near the answer.

1. i am not even sure how to begin this one. Begin by completing the square.

$
=\int_3^7 \sqrt(16t^2 + 8t + 1)dt
$

attempt:

$=\int_3^7 (16t^2 + 8t + 1)^1/2 dt$
$=\int_3^7 \frac {(16t^2 + 8t + 1)^3/2}{3/2} dt$
$=\int_3^7 \frac {2}{3} (16t^2 + 8t + 1)^3/2 dt$

2. this one i am not sure if i copied wrong or something but when i attempt it i get the answer of 1

question:
$
=\int_0^1 \frac{dx}{\sqrt(3 - 2x)}
$

attempt:

$
u = (3-2x)
$

$
du = -2dx
$

$
\frac{-1}{2}du = dx
$

$
=-\frac{1}{2}\int_0^1 u^-1/2 du = -\frac{1}{2}\int_0^1 \frac {u^1/2}{1/2} = -\frac{1}{2} * \frac{2}{1}(3-2x)^1/2
$

$
= -1 \sqrt (3-2x)
$
Looks good. Now finish.

3. this one gives me some problems as well. question says
find the equation that passes thru (1,0) is tangent to the line $6x + y = 6$ and has $y" = \frac {12}{x^3}$

answer: $\frac {6 - 6x}{x}$

4. last question says
find first quadrant area of $x + y + y^2 = 2$

answer: $\frac {7}{6}$

attempt:
first rewrite the equation to be

$x = y + y^2 - 2$

got the limits using table of values and got result of (2,1)

$A = \int_1^2 x dy$

$= \int_1^2 (y + y^2 - 2)dy$

$= \int_1^2 (\frac{y^2}{2} + \frac {y^3}{3} - 2y)$

and then i am not sure if i am on the right path because it does not work out to correct answer.What's the answer?[/quote]..
• Dec 13th 2009, 05:55 PM
TKHunny
On #4, how did you decide on the limits? Did you report the WHOLE problem statement?
• Dec 13th 2009, 06:09 PM
araym1
Quote:

Originally Posted by TKHunny
On #4, how did you decide on the limits? Did you report the WHOLE problem statement?

using a table of values, set x=0 then y=0 and yes whole problem reported and the answer to the ? is

$\frac {7}{6}$
• Dec 13th 2009, 06:41 PM
araym1
wow i can not believe i have never thought of completing the square, but that totally worked, it was the key that i was missing all along. thanks to both of you
• Dec 14th 2009, 03:46 AM
TKHunny
Quote:

Originally Posted by araym1
using a table of values, set x=0 then y=0 and yes whole problem reported and the answer to the ? is

$\frac {7}{6}$

Still insufficient. Both sections in Quadrant I are unbounded. Quadrant III would be a different story.

Ah! I see the problem. Your algebra is bad. It should be x = 2 - y - y^2. Give it another go.
• Dec 14th 2009, 06:36 AM
araym1
Quote:

Originally Posted by TKHunny
Still insufficient. Both sections in Quadrant I are unbounded. Quadrant III would be a different story.

Ah! I see the problem. Your algebra is bad. It should be x = 2 - y - y^2. Give it another go.

that also works and it changed my limits to 0,1