00:01
To evaluate the given integral, we note that this is improper since square root of x equals 0 whenever x is 0, implying that e raised to negative square root of x over the square root of x has infinite discontinuity at x equals 0.
00:17
So by definition, this is equal to the limit as c approaches 0 from the right of the integral from c to 4 of e raised to negative square root of x over the square root of x d.
00:30
Now if we apply u substitution, we would let u equal to negative square root of x, which gives us du equal to negative 1 over 2 squared of x d x or negative 2 du is equal to 1 over the squared of x d x.
00:50
So by substitution this is equal to the limit as c approaches 0 from the right of the integral from c to 4.
01:01
Of e raised to u times negative to du, that's equal to negative 2 times the limit as c approaches 0 from the right of the integral from c to 4 of e raise to u.
01:16
Now integrating this, we get negative 2 times the limit as c approaches zero from the right of e raised to u that's evaluated from c to 4.
01:26
And since u is negative square of x, then this is equal to negative 2 times the limit as c approaches 0 from the right of e raised to negative square of x, this evaluated from c to 4...