I am assuming you wish to compute:
∫∫ cos(y^2) dy dx (from y=x to 1) (from x=0 to 1).
Note that ∫ cos(y^2) dy is a non-elementary integral, so we can cannot compute it. To fix this problem, we need to switch the order of integration.
Note that you can write this integral as:
∫∫D cos(y^2) dA,
where:
D = {(x, y) | 0 <= x <= 1, x <= y <= 1}.
If you draw this region out, you can see that an alternative description of this region is:
D = {(x, y) | 0 <= x <= y, 0 <= y <= 1}.
Thus:
∫∫D cos(y^2) dA = ∫∫ cos(y^2) dx dy (from x=0 to y) (from y=0 to 1)
= ∫ y*cos(y^2) dy (from y=0 to 1) (note that this can integral is easily computed)
= (1/2)sin(y^2) (evaluated from y=0 to 1)
= (1/2)sin(1).
I hope this helps!
∫∫ cos(y^2) dy dx (from y=x to 1) (from x=0 to 1).
Note that ∫ cos(y^2) dy is a non-elementary integral, so we can cannot compute it. To fix this problem, we need to switch the order of integration.
Note that you can write this integral as:
∫∫D cos(y^2) dA,
where:
D = {(x, y) | 0 <= x <= 1, x <= y <= 1}.
If you draw this region out, you can see that an alternative description of this region is:
D = {(x, y) | 0 <= x <= y, 0 <= y <= 1}.
Thus:
∫∫D cos(y^2) dA = ∫∫ cos(y^2) dx dy (from x=0 to y) (from y=0 to 1)
= ∫ y*cos(y^2) dy (from y=0 to 1) (note that this can integral is easily computed)
= (1/2)sin(y^2) (evaluated from y=0 to 1)
= (1/2)sin(1).
I hope this helps!
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Not to sure if this is what you meant but it sounds like the order of integration must be reversed first..
If I have it correct x
Drawing a graph of the boundaries and making it x-simple rather than y simple
the limits become 0
Then by putting the integration becomes a simple substitution as it is cos(y^2)y
If I have it correct x
the limits become 0