solve the double integral of ∫∫xcosy da with a domain bounded by y=0, y=x^2, and x=1?

Question

The final answer is (1-COS1)/2 which i cannot get at all

Answer 1

I'm assuming you mean:

∫∫x cosy dy dx

where the bounds of the outer integral are x=0 to x=1 and the inner integral are y=0 to y=x².

First evaluate the inner integral from y=0 to y=x², pretending x is constant.

∫x [∫cosy dy] dx = ∫x [siny] dx = ∫x [sin(x²) - sin(0)] dy = ∫xsin(x²)dx

Then use u substitution, u = x², x = √u, du = 2x dx, so dx = du/2x = 1/(2√u) du:

∫√u sin(u) 1/(2√u) du = 1/2 ∫ sin(u) du = [-1/2 cos u] = [-1/2 cos(x²)], evaluated from x=0 to x=1

-1/2[cos1 - cos 0] = (1 - cos1)/2

Not bad for not having done calc since 1992, huh?

Answer 2

Solve Double Integral

Answer 3

The ans will be (1-COS1)/2 if the limit of x is from 0 to 1.

First integral (outer) will have limit as x = 0 to 1
Inner integral will have limit as y = 0 to x^2
da = dx.dy
First integrate wrt y
you will get siny and limits are 0 and x^2
so u r left with integral(xsin(x^2)).Limits are 0 to 1
since it can be written as integral(d(-cos(x^2))/2.
we get result as (-cos(x^2))/2 with limits 0 to 1
Substituting we get (1-cos1)/2

Answer 4

OK use da =dxdy and integrate with respect to x first at the bounds 1 and the square root of y=x this will give the second integral in terms of y only and should be easier to solve.

Answer 5

Important: you are integrating x from 0 to 1, but y from 0 to x^2. You are integrating below the parabola, not a square.


int (0 to 1) [int (0 to x^2) x cosy dy] dx =

int (0 to 1) [int (0 to x^2) cos y dy] x dx=

int (0 to 1) [sin y (0 to x^2)] x dx =

int (0 to 1) [sin (x^2) - sin 0 ] x dx =

int (0 to 1) sin (x^2) x dx

(1/2) int (0 to 1) sin (x^2) 2xdx =

1/2 int (0 to 1) sin u du = - (1/2) cos u (0 to 1)

u = x^2, du = 2xdx

u(1) = 1
u(0) = 0

So, the result is: (cos 0 - cos 1)/2 = (1 - cos 1)/2

You were right

Ana

Answer 6

good question but i don't know answer
:-(((