What is #int_0^2##int_0^1##ysqrt(y^2 + 7x)#?

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Answer 1 · 2016-04-12T22:53:37

It depends on the order in which we integrate.

#int_0^2 int_0^1 ysqrt(y^2 + 7x) dx dy ~~ 4.56514#

#int_0^2 int_0^1 ysqrt(y^2 + 7x) dy dx ~~ 206106#

Here is the integral

#int_0^2 int_0^1 ysqrt(y^2 + 7x) dx dy #

#int_0^1 ysqrt(y^2 + 7x) dx = 2/21 [y(y^2+7x)^(3/2)]_0^1# (by substitution)

# = 2/21[y(y^2+7)^(3/2)-y(y^2)^(3/2)]#

Now evaluate

#int_0^2 2/21[y(y^2+7)^(3/2)-y(y^2)^(3/2)]dy # (by substitution)

# = 2/105 (y^2+7)^(5/2)-(y^2)^(5/2)]_0^2#

# = 2/105[11^(5/2)-4^(5/2)-7^(5/2)]#

# = 2/105(121sqrt11-32-47sqrt7) ~~ 4.56514#