What is f(x) = int -9x+5sqrt(x^2+1) dxf(x)=∫−9x+5√x2+1dx if f(2) = 7 f(2)=7?

1 Answer
Noah G
Mar 4, 2017

f(x) = -9/2x^2 + 5/2secxtanx +5/2ln|secx + tanx| + 8.06f(x)=−92x2+52secxtanx+52ln|secx+tanx|+8.06

Explanation:

Separate the integrals.

f(x) = int -9xdx + int 5sqrt(x^2 +1)dxf(x)=∫−9xdx+∫5√x2+1dx

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

Since int 5sqrt(x^2 + 1)∫5√x2+1 is of the form a^2 + x^2a2+x2, we use the substitution x = tanthetax=tanθ. This means that dx = sec^2theta d thetadx=sec2θdθ.

5int sqrt(x^2 + 1)dx = 5int sqrt((tan theta)^2 + 1) * sec^2theta d theta5∫√x2+1dx=5∫√(tanθ)2+1⋅sec2θdθ

5int sqrt(x^2 + 1)dx = 5int sqrt(sec^2theta) * sec^2theta d theta5∫√x2+1dx=5∫√sec2θ⋅sec2θdθ

5int sqrt(x^2 + 1)dx = 5int sec^3theta d theta5∫√x2+1dx=5∫sec3θdθ

This is a known integral that can be found here.

5intsqrt(x^2 +1)dx = 5(1/2secxtanx + 1/2ln|secx + tanx|)5∫√x2+1dx=5(12secxtanx+12ln|secx+tanx|)

5intsqrt(x^2 + 1)dx = 5/2secxtanx + 5/2ln|secx + tanx|5∫√x2+1dx=52secxtanx+52ln|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| + Cf(x)=−92x2+52secxtanx+52ln|secx+tanx|+C

We now solve for CC.

7 = -9/2(2)^2 + 5/2sec(2)tan(2) + 5/2ln|sec2 + tan2| + C7=−92(2)2+52sec(2)tan(2)+52ln|sec2+tan2|+C

Using a calculator, we get an approximation of 8.0648~~ 8.068.0648≈8.06.

Hopefully this helps!

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