FInd numeric value of #int_1^esqrt(lnx)dx-int_0^1e^(x^2)dx#?

1 Answer
Jul 10, 2018

Answer:

#approx -0.207#

Explanation:

#I = underbrace(int_1^esqrt(lnx)dx)_(I_1) color(red)(-) underbrace( int_0^1e^(x^2)dx)_(I_2)#

For #I_1#:

Let: #y = sqrt(lnx) #

  • #implies { (x = e^(y^2)),(dx = 2 ye^(y^2) dy):}#

#:. I_1 = int_(0)^(1) \ 2y^2e^(y^2) dy#

Switching the letter of the dummy variable #y# back to #x#.

#I = I_1 color(red)(-) I_2 #

#= (I_1 + I_2) - 2 I_2#

#= int_(0)^(1) \ underbrace(2x^2e^(x^2) + e^(x^2))_(= d/dx( x e^(x^2) )) dx - 2 underbrace( int_0^1e^(x^2)dx )_(= sqrtpi/2 "erf"i(1) approx 1.463)#

#approx e - 2(1.463) approx -0.207#