# Given dy/dx=35x^6-9*e^3x find y.

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### 2 Answers

It is given that dy/dx=35x^6-9*e^3x

y = Int [ (dy/dx) dx]

=> Int [ (35x^6 - 9*e^3x ) dx]

=> Int [ 35x^6 dx] - 9*Int[e^3x dx]

=> 35*x^7/7 - 9* e^3x / 3

=> 5*x^7 - 3*e^3x + C

**The function y = 5*x^7 - 3*e^3x + C**

To find out the original, when knowing it's derivative, we'll have to integrate the expression of derivative.

We'll determine the indefinite integral of f'(x)=35x^6-9*e^3x

Int dy = y + C

Int (35x^6-9*e^3x)dx

We'll use the property of the indefinite integral, to be additive:

Int (35x^6-9*e^3x)dx =Int (35x^6)dx - Int (9e^3x)dx

Int (35x^6)dx = 35*x^(6+1)/(6+1) + C

Int (35x^6)dx = 35x^7/7 + C

Int (35x^6)dx = 5x^7 + C (1)

Int 9e^3x dx = 9*e^3x/3 + C

Int 9e^3x dx = 3e^3x + C (2)

We'll subtract (2) from (1) and we'll get:

Int (35x^6-9*e^3x)dx = 5x^7 - 3e^3x + C

**So, the primitive function is: y = 5x^7 - 3e^3x + C**