Answer:
[tex]f(x) = x^3 -sinx +Cx+D[/tex]
Step-by-step explanation:
Given that:
[tex]f ''(x)= 6x +sinx[/tex]
We are given the 2nd derivative of a function f(x) and we need to find f(x) from that.
We will have to integrate it twice to find the value of f(x).
Let us have a look at the basic formula of integration that we will use in the solution:
[tex]1.\ \int {(a\pm b)} \, dx =\int {a} \, dx + \int {b} \, dx \\2.\ \int {x^n} \, dx = \dfrac{x^{n+1}}{n+1}+C\\3.\ \int {sinx} \, dx = -cosx+C\\4.\ \int {cosx} \, dx = sinx+C[/tex]
[tex]\int\ {f''(x)} \, dx =\int\ {(6x +sinx)} \, dx \\\Rightarrow \int\ {6x} \, dx + \int\ {sinx} \, dx \\\\\Rightarrow 6\dfrac{x^2}{2} -cosx +C\\\Rightarrow 3{x^2} -cosx +C\\\Rightarrow f'(x)=3{x^2} -cosx +C\\[/tex]
Now, integrating it again to find f(x):
[tex]f(x) =\int {f'(x)} \, dx =\int{(3{x^2} -cosx +C)} \, dx \\\Rightarrow \int{3{x^2}} \, dx -\int{cosx} \, dx +\int{C} \, dx\\\Rightarrow 3\times \dfrac{x^3}{3} -sinx +Cx+D\\\Rightarrow x^3 -sinx +Cx+D\\\\\therefore f(x) = x^3 -sinx +Cx+D[/tex]
