Question

Calculus please help

Answer 1

`f(x)=(1)/(20)x^5+\sinh(x)-(1)/(2)\sinh(2)x+6`

Explanation:

Given:

`\phantom{ww} \bullet\;\;\;f''(x)=x^3+\sinh(x)`

`\phantom{ww} \bullet\;\;\;f(0)=6`

`\phantom{ww} \bullet\;\;\;f(2)=7.6`

To find f'(x), integrate f''(x):

`\begin{aligned}\displaystyle f'(x)=\int f''(x)\; \text{d}x&=\int \left(x^3+\sinh(x)\right)\; \text{d}x\\\\&=\int x^3\;\text{d}x+\int \sinh(x)\; \text{d}x\\\\&=(1)/(4)x^4+\cosh(x)+\text{K}\end{aligned}`

To find f(x), integrate f'(x):

`\begin{aligned}\displaystyle f(x)=\int f'(x)\; \text{d}x&=\int \left((1)/(4)x^4+\cosh(x)+\text{K}\right)\;\text{d}x\\\\&=\int (1)/(4)x^4\; \text{d}x+\int \cosh(x) \; \text{d}x + \int \text{K}\; \text{d}x\\\\&=(1)/(20)x^5+\sinh(x)+Kx+\text{C}\end{aligned}`

Substitute f(0) = 6 to determine the value of the constant C:

`\begin{aligned}f(0)=(1)/(20)(0)^5+\sinh(0)+K(0)+\text{C}&=6\\\\0+0+0+\text{C}&=6\\\\\text{C}&=6\end{aligned}`

Substitute f(2) = 7.6 and C = 6 to determine the value of the constant K:

`\begin{aligned}f(2)=(1)/(20)(2)^5+\sinh(2)+K(2)+6&=7.6\\\\1.6+\sinh(2)+2K+6&=7.6\\\\\sinh(2)+2K&=0\\\\2K&=-\sinh(2)\\\\K&=-(1)/(2)\sinh(2)\end{aligned}`

Therefore, function f(x) is:

`\boxed{f(x)=(1)/(20)x^5+\sinh(x)-(1)/(2)\sinh(2)x+6}`

`\textsf{As}\;\;\sinh(2)=(e^4-1)/(2e^2),\;\textsf{we can also write the equation as:}`

`f(x)=(1)/(20)x^5+\sinh(x)-(1)/(2)\left((e^4-1)/(2e^2)\right)x+6`

`f(x)=(1)/(20)x^5+\sinh(x)-\left((e^4-1)/(4e^2)\right)x+6`