Integration - areas enclosed by the y -axis?

Question

ok. another one. ive done the first bit but its not working
find the area enclosed between the curive y = (x^3) - 2 and the y-axis y = -1 and y =25
please do it step by step
thank u

60 units sq is what the answer is but how. i ahev a fair idea of what i doing, but i keep on doing something wrong and i dont know why im getting it wrong. its frustrating. grr

the penny finally dropped thanks

i was integrating inside the brackets

can u please tell me what im doing wrong. i suck at this.
heres another Q. find the area enclosed between the lines y = 4 and y = 1-x in the second quadrant. ive down a diagram and everything. im not getting the answer which is 4.5 units sq. help

Answer 1

Area = Integral of F(x)dy
y = x^3 - 2 so x = (y+2)^1/3

=[3/4. (y+2)^4/3] - now you know why 25 was chosen!

I assume you know how to evaluate this by putting in y=25 and y=-1

Answer 2

I get 60 sq units.
- - - - - - - - -

Finding the area enclosed by...

Your last 2 functions: y = -1 and y =25 are horizontal lines below and above the x-axis respectively. Your y-axis line cuts thru these horizontal lines, so we're curious as to whether the more "interesting" function

y = (x^3) - 2 crosses the y-axis (esp. in the interval
-1 < = y < = 25) because if it does then you'll have to break up your interval of integration into 2 or maybe more intervals.

So I usually start with:

integral low limit -1 to 25 [integrand] dY

and then if I find it necessary, I'll break the integral into more intervals as needed. That done, I observe the integrand must be a function of "y" (not "x") to be manageable; so I take the function I called more "interesting" and trigger with it:

x = (y + 2)^ (1/3) noticing that in the interval -1 < = y < = 25
x stays right of the y-axis (i.e. x > 0 for the interval above); therefore I don't have to break up the interval. So your integrand is

(y + 2)^ (1/3) which after integration is (3/4) [y + 2]^(4/3) to be evaluated at the limits.

Answer 3

The above are all good answers. If you don't think about inverting the function, here goes...

The curve crosses y=-1 at x=1.
It crosses y=25 at x=3.

You need to split your area in two:
1) The rectangle delimited by y=25, y=-1, x=0 and x=1. It's easy to figure out the area for this part.

2) The area delimited between the curves y=25 and y=x³-2, from x=1 to x=3. The area is
integral (1 to 3) (25 - (x³-2)) dx
= integral (1 to 3) (27 - x³) dx.

(1) and (2) will add up to 60. Have fun.

Answer 4

Dear
First I want to make one point clear
Finite integration give the area under the curve bounded by the X-Axis and the limit of integration

one Soln and it work as a dream
Just find the inverse Fn
X = (Y+2)^ 1/3
Then Integrate X with respect to Y from -1 to 25
This will solve all
And Remember that I am the First to say that

Answer 5

y = x³ - 2
x³ = y + 2
x = (y + 2)^(1/3)
Area = A
A = ∫(y + 2)^(1/3) dy between given limits
A = (y + 2)^(4/3) / (4/3)
A = (3/4).(y + 2)^(4/3)
Inserting limits:-
A = (3/4) [ 27^(4/3) - 1 ]
A = (3/4).[ 81 - 1 ]
A = 60

Answer 6

we can restate the question through noting that x = a million/sqrt(y) Then, we merely opt to discover the crucial from a million to 4 of one million/sqrt(y) considering that a million/sqrt(y) = y^(-a million/2), that is crucial is y^(a million/2)/(a million/2) = 2sqrt(y) comparing at y=4 and y=a million, we get 2sqrt(4) - 2sqrt(a million) = 4 - 2 = 2