Respuesta :
Answer:
[tex]f(x)=\frac{x^3}{3}+9e^x+Cx+D[/tex]
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).
[tex]\int {f(x)} \, dx =F(x)+C[/tex]
To find the first antiderivative you must integrate the function [tex]f''(x) = 2x + 9e^x[/tex]
[tex]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[/tex]
To find the second antiderivative you must integrate the function [tex]f'(x) =x^2+9e^x+C[/tex]
[tex]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[/tex]
Therefore,
[tex]f(x)=\frac{x^3}{3}+9e^x+Cx+D[/tex]
The function "f" will be "[tex]\frac{x^3}{3}+9e^x+Cx+D[/tex]".
Given function,
- [tex]f''(x) = 2x+9e^x[/tex]
By integrating, we get
→ [tex]f'(x) = \int f'' (x) dx[/tex]
[tex]= \int (2x+9e^2)dx[/tex]
[tex]= x^2+9e^x+C[/tex]
Again by integrating with respect to "x", we get
→ [tex]f(x) = \int f'(x) dx[/tex]
[tex]= \int (x^2+9e^x+C)dx[/tex]
[tex]= \frac{x^3}{3}+9e^x+Cx+D[/tex]
Thus the above answer is right.
Learn more:
https://brainly.com/question/23084452
