00:01
In this problem we are asked to evaluate the double integral over the region r of 2 times x times y d x, dx, d .y.
00:13
Here we are given that the region r is bounded by the curves y equals to x squared plus 4, y equals to 3 times x plus 2, x equals to negative 2 and x equals to 1.
00:29
So let us substitute these for the limits of the integral.
00:34
We have integral negative 2 to 1, integral x squared plus 4 up to 3 times x plus 2, 2 times x times y, d.
00:48
D .x.
00:50
Let us evaluate the innermost integral.
00:53
We get integral negative 2 to 1, 2 times x times the integral of y is 1.
01:01
Y squared over 2 with lower limit x squared plus 4 and upper limit 3 times x plus 2 so here 2 and 2 would get cancelled now let us substitute the limits we get integral negative 2 to 1 x times substituting the upper limit we have 3 times x plus 2 raised to the part 2 minus substituting the lower limit we have x squared plus 4 the whole squared d x.
01:31
Let us simplify the integrant.
01:35
We have 3 times x plus 2 the whole square can be expanded using the formula for a plus b the whole squared.
01:46
So we have 9x squared plus 4 plus 3 times 2 times 2 which is 12 times x minus of again making use of the same formula we have x raised to the power 4 plus 6.
02:03
Plus 4 times 2 which is 8 so 8 times x squared d x.
02:12
Further simplifying we have integral negative 2 to 1 x times 9 x squared plus 4 plus 12 times x minus x raised to the power 4 minus 16 minus 8 times x squared d x so here let us combine the like terms.
02:41
We get x squared minus 12 plus 12 times x minus x raised to the power 4 d x.
02:52
Next let us multiply the x inside with all the four terms.
03:02
So doing so we obtain integral negative 2 to 1 x cubed minus 12 times x plus plus 12 times x squared minus x raised to the power 5.
03:24
Now we can evaluate the integral...