If #g(x)# is the antiderivative of #f(x) = root3(x^2+4x)#, what is the value of #g(1)#?

#g(5) = 7#.

1 Answer
Jim H
Jan 6, 2017

#g(1) ~~ -3.882#

Explanation:

The Fundamental Theorem of Calculus tells us that

#int_1^5 root3(x^2+4x) dx = g(5)-g(1)#

Approximation or technology gets us

#int_1^5 root3(x^2+4x) dx ~`~ 10.882#.

So

#10.882 ~~ 7 - g(1)# and #g(1) ~~ 7-10.882 = -3.882#.

You'll probably need approximation or technology because the general antiderivative involves the hypergeometric function which is not taught in introductory calculus classes.

#int root3(x^2+4x) dx = (3root(3)(x^2+4x)(8((x+4)/x)^(2/3)color(white)(.)_2F_1(1/3,2/3;4/3;-4/x)+x^2+6x+8))/(5(x+4))+C#

where #color(white)(.)_2F_1(a,b;c;x)# is the hypergeometric function. (No, I cannot explain it.)

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