How to find the indefinite integral

Find a function y = f(x) that satisfies both conditions:

dy/dx = 3x^2 - 2 ; f(0) = 4

Integrate dy/dx so you get

y = x^3 - 2x + C

f(0) = 4 so

4 = (0)^3 - 2(0) + C
C = 4

therefore f(x) = x^3 - 2x + 4

dy/dx = 3x^(2) - 2 ; f(0) = 4

Multiply both sides of the equation by dx.

dy = [ 3x^(2) - 2 ] dx

Use the separation of variables technique and integrate both sides. You can take the constants outside the integrals.

integral [dy] = 3 * integral [x^(2) dx] - 2 * integral [dx]

Integrate using the power rule of antidifferentiation method:
integral [x^(n) dx] = [x^(n+1)] / [(n + 1) ] + C, n doesn't equal - 1.


y = 3 * [(1/3) * x^(3) ] - 2x + C

Multiply 3 times (1/3) = 1.

y = x^(3) - 2x + C

Given the initial condition: f(0) = 4 where x = 0 and y = 4.

Plug these values here and solve for C.

4 = (0)^(3) - 2(0) + C

C = 4 ---> Substitute the constant of integration back into the y function.

y = x^(3) - 2x + 4 -----------> ANSWER

the integral of 3x^2 - 2 is:: x^3 - 2x because if you differentiate x^3 you get 3x^2 and if you differeniate -2x you get -2

so the integral equals x^3 - 2x + C, the constant

you know that when x equals zero, f(x) equal 4

so 0^3 - 2(0) + C = 4

so C equals 4

x^3 - 2x + 4