# `f'(x) = sinh(x) + 2cosh(x), f(0) = 2` Find `f`.

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### 2 Answers

You need to evaluate the function f using the provided information, hence, you need to apply the antiderivative, such that:

`int (f'(x) sinhx + 2cosh x)dx = int (f'(x)(e^x - e^(-x))/2) + (e^x + e^(-x)))dx`

`int (f'(x) sinhx + 2cosh x)dx = (1/2)int (f'(x)(e^x)dx - (1/2)int f'(x) e^(-x)dx + int e^x dx + int e^(-x)dx`

You need to evaluate each of the two integrals separately. You may integrate `int (f'(x)(e^x)dx` using integration by parts. You may consider u = e^x and dv = f'(x), such that:

`int udv = uv - int vdu`

`u = e^x => du = e^x`

`dv = f'(x) => v = f(x)`

`int f'(x) e^x dx = f(x) e^x - int f(x)e^x => (1/2)int f'(x) e^x dx = (1/2)f(x) e^x - (1/2)int f(x)e^x`

`(1/2)int f'(x) e^(-x)dx = (1/2)f(x)e^(-x) + (1/2)int e(-x)f(x) dx `

`u = e^(-x) => du = -e^(-x)`

`dv = f'(x) => v = f(x)`

`(1/2)int (f'(x)(e^x)dx - (1/2)int f'(x) e^(-x)dx= (1/2)f(x) e^x - (1/2)int f(x)e^x - (1/2)f(x)e^(-x) - (1/2)int e(-x)f(x) dx`

`int e^x dx + int e^(-x)dx = e^x - e^(-x) + c`

Hence, evaluating the integral, yields:

`int (f'(x) sinhx + 2cosh x)dx = (1/2)f(x) e^x - (1/2)int f(x)e^x - (1/2)f(x)e^(-x) - (1/2)int e^(-x)f(x) dx + e^x - e^(-x) + c`

`int (f'(x) sinhx + 2cosh x)dx = (1/2)f(x)(e^x - e^(-x)) - (1/2)int (f(x)e^x -e(-x)f(x))dx + e^x - e^(-x) + c`

`int (f'(x) sinhx + 2cosh x)dx = f(x)sinh x - int f(x) sinh x + 2sinhx + c`

**Given the provided information, this is the answer for the antiderivative `int (f'(x) sinhx + 2cosh x)dx = f(x)sinh x - int f(x) sinh x + 2sinhx + c.` **

`f'(x) = sinhx+2coshx`

Integrating both sides with respect to x ,w get

`intf'(x)=int(sinhx+2coshx)dx`

`f(x) = int((e^x-e^(-x))/2 +2(e^x+e^(-x))/2)dx`

`=int((3e^x+e^(-x))/2)dx`

`=(3e^x-e^(-x))/2+c`

Given `f(0) = 2`

`f(0) = 1+c`

`c = 1`

Therefore` f(x) = (3e^x-e^(-x))/2+1`