# Thread: Integration qs and checks

1. ## Integration qs and checks

Here a few questions and my answers where i have done. check means i want a yes or no as to my answer =P

Here goes:

Q1: 2xy(y')=y^2-9 x not=0

b) find the general solution:

y'=(y^2-9)/(2xy) ? check

c) Find the particular solution when y=5 and x=4

Q2: The half life of morphine in the human bloodstream is 3 hours.

(a) If initially there is 0.4 mg of morphine in the bloodstream, find an equation for the amount of morphine in the bloodstream at any time t.

M=morphine amount, t=time in hours

M=(cuberoot(0.5)^t)*)0.4 ? check

(b) When does the amount drop below 0.01 mg?

15.96 hours? check

Q3: Find a function f(x) such that the point (1, 2) is on the graph of y=f(x), the slope of the tangent line at (1, 2) is 3 and f''(x)=x-1

I got f(x)=(x^3)/6-(x^2)/2+3.5x-1/1/6 check please...(1/1/6 means 1 and 1 sixth)

2. Originally Posted by deragon999
Here a few questions and my answers where i have done. check means i want a yes or no as to my answer =P

Here goes:

Q1: 2xy(y')=y^2-9 x not=0

b) find the general solution:

y'=(y^2-9)/(2xy) ? check

c) Find the particular solution when y=5 and x=4

Q2: The half life of morphine in the human bloodstream is 3 hours.

(a) If initially there is 0.4 mg of morphine in the bloodstream, find an equation for the amount of morphine in the bloodstream at any time t.

M=morphine amount, t=time in hours

M=(cuberoot(0.5)^t)*)0.4 ? check

(b) When does the amount drop below 0.01 mg?

15.96 hours? check

Q3: Find a function f(x) such that the point (1, 2) is on the graph of y=f(x), the slope of the tangent line at (1, 2) is 3 and f''(x)=x-1

I got f(x)=(x^3)/6-(x^2)/2+3.5x-1/1/6 check please...(1/1/6 means 1 and 1 sixth)

$f''(x)=x-1\Rightarrow{f'(x)=\frac{x^2}{2}-x+C}$

We know that [tex]f'(1)=3

So

$3=\frac{1}{2}-1+C\Rightarrow{C=\frac{7}{2}}$

So now

$f'(x)=\frac{x^2}{2}-x+\frac{7}{2}\Rightarrow{f(x)=\frac{x^3}{6}-\frac{x^2}{2}-\frac{7x}{2}+C}$

we know $f(1)=2$

So

$2=\frac{1}{6}-\frac{1}{2}+\frac{7}{2}+C\Rightarrow{C=\frac{-5}{3}}$

$\therefore{f(x)=\frac{x^3}{6}-\frac{x^2}{2}+\frac{7x}{2}-\frac{5}{3}}$

3. are you sure about you working for C in f(1)=2, and there were others(hehe work slave work)

4. Originally Posted by deragon999
are you sure about you working for C in f(1)=2, and there were others(hehe work slave work)
You are right, it should be -7/6...well I am going to bed. I am sure someone else will help you.

Good luck

5. OK thanks for all your help.

Anyone else want to step in and look at Q1 and Q2?

6. Originally Posted by deragon999
Here a few questions and my answers where i have done. check means i want a yes or no as to my answer =P

Here goes:

Q1: 2xy(y')=y^2-9 x not=0

$2xy\frac{dy}{dx}=y^2-9 \iff \frac{y}{y^2-9}dy=\frac{1}{2x}dx$
$\int \frac{y}{y^2-9}dy= \int \frac{1}{2x}dx$
$\frac{1}{2}\ln|y^2-9|=\frac{1}{2} \ln|x|+C$
$y^2-9 = Ae^{x} \iff y=\sqrt{Ae^x+9}$