Question #37d26
2 Answers
Explanation:
Let
then
when
Now use property 2
Now use property 1
Now putting the limits we get
Explanation:
Here's another method. We call the integral
Complete the square under the
#I = int_0^2 sqrt(4(y^2 - 2y) + 5) dy#
#I = int_0^2 sqrt(4(y^2 - 2y + 1 - 1) + 5) dy#
#I = int_0^2 sqrt(4(y^2 - 2y + 1) - 4 + 5) dy#
#I =int_0^2 sqrt(4(y - 1)^2 + 1) dy#
We now let
#I = int_(-1)^1 sqrt(4u^2 +1) du#
We now make the substitution
#I = int_(-1)^1 sqrt(4(1/2tantheta)^2 + 1) * 1/2sec^2theta d theta#
#I=int_(-1)^1 sqrt(tan^2theta + 1) * 1/2sec^2theta d theta#
#I = int_(-1)^1 sqrt(sec^2theta) * 1/2sec^2theta d theta#
#I = int_(-1)^1 1/2sec^3theta d theta#
This is a known integral that is derived here
#I = [1/4secthetatantheta + 1/4ln|sectheta + tantheta|]_(-1)^1#
We return to
From our trig substitution, we obtain that
This means that
#I = [1/4sqrt(4u^2 + 1)(2u) + 1/4ln|sqrt(4u^2 + 1) + 2u|]_(-1)^1#
#I = 1/2sqrt(5) + 1/4ln|sqrt(5) + 2| - (-1/2sqrt(5) + 1/4ln|sqrt(5) - 2|)#
#I = 1/2sqrt(5) + 1/4ln|sqrt(5) + 2| + 1/2sqrt(5) - 1/4ln|sqrt(5) - 2|#
#I = sqrt(5) + 1/4ln|(sqrt(5) + 2)/(sqrt(5) - 2)|#
Hopefully this helps!