Integrate #int_1^2x\sqrt{x-1}dx# ?

2 Answers
Rhys
Jun 21, 2018

# = 16/15 #

Explanation:

let #u = x-1 #

#du = dx#

#x=2 => u = 1 #

#x = 1 => u = 0 #

Rearrange: # x = u+1 #

#=> int_0 ^1 (u+1) sqrt(u) du #

# = int_0 ^1 (u^(3/2) + u^(1/2)) du #

# = [ 2/5 u^(5/2) + 2/3 u^(3/2) ]_0 ^1 #

# = 2/5 + 2/3 = color(red)(16/15 #

Skewd
Jun 21, 2018

Use substitution, let #w=x-1# then #dw=dx# and the integral becomes #int_0^1(w+1)sqrt(w )dw#
answer: #(16/15)#

Explanation:

If you use this substitution the limits of integration are found by plugging #x=1# into the substitution #w=x-1=1-1=0# and #w=2-1=1#
x is found by solving #w=x-1# for x. #x=w+1#
Simplify the integrand algebraically
#int_0^1[wsqrt(w)+sqrt(w)]dw#
#int_0^1[w^(3/2)+w^(1/2)]dw#
Now integrate using the power rule.
#=(2/5)w^(5/2)+(2/3)w^(3/2)# from 0 to 1
#=(2/5)+(2/3)=(16/15)#

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