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.

$\displaystyle

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

$

answer =84

attempt:

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

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

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

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

$\displaystyle u = 16t^2 + 8t + 1 $

$\displaystyle du = 32t + 8 dt $

so i have tried this then...

$\displaystyle \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:

$\displaystyle

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

$

answer =0.73

attempt:

$\displaystyle

u = (3-2x)

$

$\displaystyle

du = -2dx

$

$\displaystyle

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

$

$\displaystyle

=-\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

$

$\displaystyle

= -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 $\displaystyle 6x + y = 6 $ and has $\displaystyle y" = \frac {12}{x^3} $

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

4. last question says

find first quadrant area of $\displaystyle x + y + y^2 = 2 $

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

attempt:

first rewrite the equation to be

$\displaystyle x = y + y^2 - 2 $

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

$\displaystyle A = \int_1^2 x dy $

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

$\displaystyle = \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.