Question

F(x) = integrate (t ^ 3 + 4t ^ 2 + 6) dt from 0 to x; f ^ 1 * (x) = then

Answer 1

Step 1

Write the function

`f(x)=\int_0^x\left(t^3+4t^2+6\right)dt\:`

Step 2

`\begin{gathered} \mathrm{Apply\:the\:Sum\:Rule}: \\ \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx \\ \\ ((t^4)/(4)+(4t^3)/(3)+6t)_0^x \\ \\ =(x^4)/(4)+(4x^3)/(3)+6x \end{gathered}`

Step 3

`\begin{gathered} \\ f^(\prime)(x)=(d)/(dx)\left((x^4)/(4)+(4x^3)/(3)+6x\right) \\ \\ f^(\prime)(x)=x^3+4x^2+6 \\ \\ f^(\prime)^(\prime)(x)=(d)/(dx)\left(x^3+4x^2+6\right) \\ \\ f^(\prime)^(\prime)(x)=3x^2+8x \end{gathered}`

Final answer

`f^(\prime)^(\prime)(x)=3x^2+8x`