00:01
In this question, we are asked to calculate the definite interval from 0 to 1 of the square root of 2x plus 1 dx.
00:06
And we will use u substitution.
00:09
And we will take u to be equal to 2x plus 1.
00:13
Then du equals to 2 times d x.
00:18
We also need to change the limits of iteration.
00:21
To get the new upper limit of iteration, we need to take the old upper limit of iteration and plug in for x in the expression for u.
00:29
To get 3.
00:32
Similarly, to get the new lower limit of integration, we need to take the old lower limit and plug in for x in the expression for you to get 1.
00:44
Now we can rewrite the integral as the integral from 1 to 3 of the square root of u, and d x becomes du over 2 from this equation here.
01:01
This equals to 1 1 1⁄ multiplied by the integral of u to the 1 1⁄2, because the square root is equivalent to the one -half power...