Integration question?

Question

Find the area between the curves y=2x^(2)+1 and y=13-2x

Where do I start? Can someone please explain how I would solve this using integration?

Thanks in advance.

Answer 1

Intersect when:-
2x² + 1 = 13 - 2x
2x² + 2x - 12 = 0
x² + x - 6 = 0
(x + 3).(x - 2) = 0
x = - 3, x = 2
A = ∫ (13 - 2x) - 2x² - 1 dx between -3 and 2
A = ∫ 12 - 2x - 2x² dx between lims.
A = 12x - x² - 2x³/3 between lims.
A = (24 - 4 - 16/3) - (-36 - 9 + 54/3)
A = 44/3 + 27
A = 44/3 + 81/3
A = 125/3
A = 41 (2/3) units ²

Answer 2

A quick start -- no math -- just ideas

There are at least 2 ways to read this problem
1 find the area under each curve and subtract one from the other -- this will provide an area that is a functon of x. I would not select this approach first.
2. The two curves will intersect at two values of x. (x1,x2) Between these intersections there will be an upper curve and a lower curve. The area under each curve between the values of x you found will be a constant. Subtract the smaller area from the larger area.

Calculus is used to integrate each curve between the limits x1 and x2 to find the area under it.

You can actually do it in one step by subtracting the formulas and integrating the difference between limits, but that doesn't give you the picture you asked for.

Answer 3

y = 2x^2 + 1 is a parabola that has a vertex (lowest point) one unit above the x-axis. The line
y = 13 - 2x intersects the parabola above the x axis at x = -3 and x = 2.

We use INT for the integral symbol.

INT [13 - 2x - (2x^2 + 1)] from x = -3 to x = 2.

= INT [13 - 2x - 2x^2 - 1] from x = -3 to x = 2

= INT [-2x^2 - 2x +12] from x = -3 to x = 2

= [(-2/3)x^3 - x^2 + 12x] from x = -3 to x = 2

= (-2/3)(2^3) - 2^2 +12(2) - [(-2/3)((-3)^3) - (-3)^2 + 12(-3)]

= -16/3 - 4 + 24 - [18 - 9 - 36]

= -16/3 - 4 + 24 - 18 + 9 + 36

=125/3

Answer 4

Integrate (13-2x)-(2x^2+1).
You get the lower limit and the upper limit of the integration by solving both y to be equal : 2x^2+1=13-2x. Those are the points where the graphs intersect.
Try to graph it.

Answer 5

Find the intersections for limits of integration:
y = 2x^2 + 1 = 13 - 2x
2x^2 + 2x - 12 = 0
x^2 + x - 6 = 0
(x + 3)(x - 2) = 0
x = (-3,2)
. . . . 2
A = ∫(12 - 2x - 2x^2)dx
. . -3
A = 12(2 + 3) - (4 - 9) + (2/3)(8 - 27)
A = 24 + 36 - 4 + 9 + 16/3 - 18
A = 47 + 16/3
A ≈ 52.333

Answer 6

Graph the curves on a coordinate axis. Set the two equations equal to each other to find where the graphs intersect, this will give you your limits of integration. Using the graph, see which curve is on top and which is on the bottom. Take the integral of the top function minus the bottom.

Answer 7

Start by subtracting
g(x) - f(x) = 2x^2 + 1 - (13 - 2x)
= 2x^2 + 2x -12

Then calculate where it intersects by factoring.
2 * (x + 3)(x - 2)
x = -3 or 2

So integrate the function from x = -3 to 2.

Answer 8

You have to solve them simultaneously to find where they meet. These will be your integration limits.

Work out the areas under each of them individually using integration.

Subtract the smaller answer from the larger to find the area between them.

EDIT: WillowTree suggests subtract first, then integrate. That's fine too.

Answer 9

First, find where they intersect. Those are your points of integration

Second, draw a rough sketch of your curves. The equation that you're integrating is the top curve minus your bottom curve.

Third - integrate! ^-^

Answer 10

Integration by way of factors. The functionality is a fabricated from 2 applications: (x) * (cosx) call certainly one of them "u" and the different "dv" choose for the "u" so as that its by-product simplifies u = x du = a million*dx dv = cosx v = sinx it is for the formulation setup, that's u*v - quintessential (vdu) =xsinx - quintessential(sinx * a million * dx) =xsinx - (-cosx) + C = xsinx + cosx + C