What is #f(x) = int -9x+5sqrt(x^2+1) dx# if #f(2) = 7 #?

Separate the integrals.

#f(x) = int -9xdx + int 5sqrt(x^2 +1)dx#

The first integral is easy, we can do this using #int x^n dx = x^(n + 1)/(n + 1) + C#, where #n != -1#. The second integral, though, will require trig substitution.

Since #int 5sqrt(x^2 + 1)# is of the form #a^2 + x^2#, we use the substitution #x = tantheta#. This means that #dx = sec^2theta d theta#.

#5int sqrt(x^2 + 1)dx = 5int sqrt((tan theta)^2 + 1) * sec^2theta d theta#

#5int sqrt(x^2 + 1)dx = 5int sqrt(sec^2theta) * sec^2theta d theta#

#5int sqrt(x^2 + 1)dx = 5int sec^3theta d theta#

This is a known integral that can be found here.

#5intsqrt(x^2 +1)dx = 5(1/2secxtanx + 1/2ln|secx + tanx|)#

#5intsqrt(x^2 + 1)dx = 5/2secxtanx + 5/2ln|secx + tanx|#

We now put all of this together and add the constant of integration.

#f(x) = -9/2x^2 + 5/2secxtanx +5/2ln|secx + tanx| + C#

We now solve for #C#.

#7 = -9/2(2)^2 + 5/2sec(2)tan(2) + 5/2ln|sec2 + tan2| + C#

Using a calculator, we get an approximation of #8.0648~~ 8.06#.

Hopefully this helps!