00:01
Hi here for the given question we are given integral.
00:04
We need to solve them.
00:05
The first one is integration over 0 to 2 4 x e to the power minus x square upon 2 dx.
00:13
So here in our case now for the first part of the question, we will use the u substitution method.
00:19
So here in our case, let x is equal to 2 then here we have u is equal to 2 square upon 2 which is equal to 2 and for them here we are assuming that we take x is equal to u is equal to x square upon 2.
00:49
So here now further on differentiating this value.
00:52
We have du equals to x dx.
00:57
So here we have 4 times integration over 0 to 2 e to the power minus u du.
01:03
So here now integrating this again.
01:05
We have 4 times minus e to the power minus u limit 0 to 2.
01:09
So here now substituting the values and rearranging the solution.
01:14
We have value equals to 4 into 1 minus e to the power minus 2...