I need to find the area of the region between the parabpla y^2 = 16x an the line 4x-3y=4.
I understand that i need to get them to both equal y but when i do that i seem to confuse myself and get really odd values.
Certainly we don't obtain rational solutions ( it appears $\displaystyle \sqrt{13}$ ) but, what is the problem?
Fernando Revilla
In my text book it shows that i should have the curves equal to y. So that I can find the x intercepts and work with the integral from there. However I can seem to figure out a solution for y^2=16x and 4x-3y=4. When I tried i got y=4 sqrt x and y=-(4(-x-1)/3) but i dont think thats right.
If you're studying calculus you're expected to know how to solve simultaneous equations.
I suggest finding the area between x = y^2/16 and x = (4 + 3y)/4, in which case you need the y-coordinates of the intersection points of the two curves.
Equate: y^2/16 = (4 + 3y)/4. Solve for y.
No.
Your values of y are correct (to 4dp - but are you meant to use exact values?)
Using your values of y as the integral terminals, I don't get your answer (but since you only give a final answer, it's impossible to know whether your error is in the calculus, algebra or arithmetic).
$\displaystyle \int_{-1.2111}^{13.2111}\left[(\frac{4+3y}{4}\right)-(\frac{y^2}{16})] dx$
$\displaystyle [(\frac{4y+(3y^2/2)}{4y})-(\frac{(y^3/3)}{16y})]_{-1.2111}^{13.2111}$
$\displaystyle (\frac{4(13.2111)+(3(13.2111)^2/2)}{4(13.2111)})-(\frac{(13.2111)^3/3)}{16(13.2111)})-(\frac{4(-1.2111)+(3(-1.2111)^2/2)}{4(-1.2111)})-(\frac{(-1.2111)^3/3)}{16(-1.2111)})$
I had another go the way i was intending to do it in the first place. and got a result of 1.742.
The last way I tried to combine the 2 fractions to make it a bit easier it looked like this
$\displaystyle \int_{-1.2111}^{13.2111}\left(\frac{4+3y-y^2}{4-16}\right) dx$
But i think i got that wrong
You must have learnt the rule $\displaystyle \displaystyle \int a y^n \, dy = \frac{a}{n+1} y^{n+1}$ (ignoring the arbitrary constant of integration), $\displaystyle n \neq -1$.
Both of your integrals are of this type. There is no point doing questions requiring the application of integration if you are struggling with things like this.
No. The first integral is wrong. Do you realise that $\displaystyle \displaystyle \frac{4 + 3y}{4} = 1 + \frac{3}{4} y$ and so the integral is .....
You are strongly advised to review basic integration techniques (and the pre-requisite algebra) before attempting questions involving applications of integration.