antiderivative
f''=3x+5sin(x)
f(0)=4
f'(0)=4
Find f(x)
I got (1/2)x^3-5sin(x)+(15/2)x+(7/2), but it was wrong
To find the antiderivative of f''(x) = 3x + 5sin(x), we integrate term by term.
∫(f''(x) dx) = ∫(3x + 5sin(x) dx)
First, let's find the antiderivative of 3x:
∫(3x dx) = (3/2)x^2 + C1
Next, let's find the antiderivative of 5sin(x):
∫(5sin(x) dx) = -5cos(x) + C2
Now we have:
f'(x) = (3/2)x^2 - 5cos(x) + C1 + C2
Given f'(0) = 4, we substitute x = 0:
4 = (3/2)(0^2) - 5cos(0) + C1 + C2
Simplifying, we get:
4 = 0 - 5 + C1 + C2
9 = C1 + C2
Next, let's find the antiderivative of f'(x):
∫(f'(x) dx) = ∫((3/2)x^2 - 5cos(x) + C1 + C2) dx
Using the power rule, we integrate the first term:
(3/2) ∫(x^2 dx) = (3/2)(1/3)x^3 + C3
Next, we integrate the second term:
-5 ∫(cos(x) dx) = -5sin(x) + C4
Adding the constants C1 + C2, we have:
f(x) = (3/2)(1/3)x^3 + C3 - 5sin(x) + C4
Now, using the given condition f(0) = 4, we substitute x = 0:
4 = (3/2)(1/3)(0^3) + C3 - 5sin(0) + C4
Simplifying, we get:
4 = 0 + C3 - 0 + C4
4 = C3 + C4
Now, combining the constants C3 and C4:
f(x) = (1/2)x^3 - 5sin(x) + C
Therefore, the correct antiderivative f(x) is:
f(x) = (1/2)x^3 - 5sin(x) + C
Remember, the constant C represents any arbitrary constant that can take any value.