Question

Find f. (Use C for the constant of the first antiderivative and D for the constant of the second antiderivative. f"(x) = 2x + 9ex Rx) =

Answer 1

Answer:

$f(x)=\frac{x^3}{3}+9e^x+Cx+D$

Step-by-step explanation:

The function F(x) is an antiderivative of the function f(x) on an interval I if

F′(x) = f(x) for all x in I.

The function F(x) + C is the General Antiderivative of the function f(x) on an interval I if F′(x) = f(x) for all x in I and C is an arbitrary constant.

The Indefinite Integral of f(x) is the General Antiderivative of f(x).

$\int {f(x)} \, dx =F(x)+C$

To find the first antiderivative you must integrate the function $f''(x) = 2x + 9e^x$

$f'(x)=\int { 2x + 9e^x} \, dx \\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int \:2xdx+\int \:9e^xdx = x^2+9e^x\\\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\\\x^2+9e^x+C$

To find the second antiderivative you must integrate the function $f'(x) =x^2+9e^x+C$

$f(x)=\int {x^2+9e^x+C} \, dx\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int \:x^2dx+\int \:9e^xdx+\int \:Cdx= \frac{x^3}{3}+9e^x+Cx\\\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\\\\frac{x^3}{3}+9e^x+Cx+D$

Therefore,

$f(x)=\frac{x^3}{3}+9e^x+Cx+D$