Question #37d26

2 Answers
Apr 8, 2017

#int_0^2sqrt(4y^2-8y+5) dy=sqrt5+1/2log(sqrt5+2)~~2.957#

Explanation:

#intsqrt(x^2+a^2)=x/2sqrt(x^2+a^2)+a^2/2log (x+sqrt(x^2+a^2))#.......Property 1
#int_a^bf(x)dx=int_a^bf(a+b-x)dx# ........Property 2

#int_0^2sqrt(4y^2-8y+5) dy=2int_0^2sqrt(y^2-2y+5/4)dy#
#rArrint_0^2sqrt(4y^2-8y+5) dy=2int_0^2sqrt((y-1)^2+(1/2)^2#
Let #y-1=t#
then #dy=dt#
when #y=0 ,t=-1#
#when y=2,t=1#

#rArrint_0^2sqrt(4y^2-8y+5) dy=2int_-1^1sqrt(t^2+(1/2)^2)#
Now use property 2

#2int_-1^1sqrt(t^2+(1/2)^2)=2xx2int_0^1sqrt(t^2+(1/2)^2)#
Now use property 1
#=4xxt/2sqrt(t^2+(1/2)^2)+4xx1/2xx(1/2)^2log(t+sqrt(t^2+(1/2)^2))#
Now putting the limits we get

#2xx2int_0^1sqrt(t^2+(1/2)^2)=[2xxsqrt5/2+1/2xxlog(1+sqrt5/2)]-[1/2xxlog(1/2)]#
#=sqrt5+1/2log(sqrt5+2)#

Apr 8, 2017

#I = sqrt(5) + 1/4ln|(sqrt(5) + 2)/(sqrt(5) - 2)|#

Explanation:

Here's another method. We call the integral #I#.

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 #u = y - 1#. Then #du = dy#.

#I = int_(-1)^1 sqrt(4u^2 +1) du#

We now make the substitution #u = 1/2tantheta#. Then it follows that #du = 1/2sec^2theta d theta#.

#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 #u#, where we can evaluate the integral (I didn't change the bounds of integration, so the answer that you would get from evaluating in #theta# would be wrong).

From our trig substitution, we obtain that #2u = tantheta#. So, the side opposite #theta# measures #2u# and the side adjacent measures #1#. Then the hypotenuse would measure #sqrt(4u^2 + 1)#.

This means that #sectheta = sqrt(4u^2 + 1)/1 = sqrt(4u^2 + 1)#.

#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!