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

1 Answer
Mar 4, 2017

f(x) = -9/2x^2 + 5/2secxtanx +5/2ln|secx + tanx| + 8.06

Explanation:

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!