How do you evaluate the definite integral #int (4x)/sqrt(x^2+1) dx# from # [-sqrt3, sqrt3]#?

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Answer 1 · 2017-01-13T14:35:37

#int_(-sqrt(3))^sqrt(3) (4x)/sqrt(x^2+1) dx = 0#

#int_(-sqrt(3))^sqrt(3) (4x)/sqrt(x^2+1) dx = 2int_(-sqrt(3))^sqrt(3) (d(x^2+1))/sqrt(x^2+1) dx = 4 [sqrt(x^2+1)]_(-sqrt(3))^sqrt(3)=4(2-2) = 0#

We might have spared ourselves the trouble of doing the calculations, noting that:

#f(x) = (4x)/sqrt(x^2+1)#

is an odd function, that is:

#f(-x) = -f(x)#

If such a function can be integrated over the interval #(-a,a)# the integral is always null as:

#int_(-a)^a f(x)dx = int_(-a)^0 f(x)dx + int_0^a f(x)dx #

Substitute #t=-x# in the first integral:

#int_(-a)^0 f(x)dx = int_a^0 f(-t)d(-t) = -int_0^a f(-t)d(-t) = int_0^a f(-t)dt = - int_0^a f(t)dt#

So we get:

#int_(-a)^a f(x)dx = int_0^a f(x)dx - int_0^a f(x)dx = 0#