If f(x)= (2x+7)^(1/2), where F'(X)=f(x) and F(1)=7, then F(x)=?
I know I need to use integration but I do not know how to take the anti derivative of the equation
F ' (x) = (2x+7)^(1/2)
F(x) = (2/3)(1/2) (2x+7)^(3/2) + c
F(1) = 7
7 = (2/3)(1/2)(9^(3/2)) + c
7 = (1/3)( 27) + c
7 = 9 + c
c = -2
you finish it
where does the one half come from in F(x)
d/dx (2x+7)) = 2
so the 1/2 gets rid of the 2
ok thanks
To find the antiderivative (or integral) of a function, you need to apply the reverse operation of taking the derivative. In this case, you need to find the antiderivative of f(x)=(2x+7)^(1/2) with respect to x.
To do this, you can apply a basic integration rule called the Power Rule:
∫x^n dx = (x^(n+1))/(n+1) + C
where n is any real number and C represents the constant of integration.
In the given equation, f(x)=(2x+7)^(1/2), the exponent is 1/2. Applying the Power Rule with n = 1/2, we have:
∫(2x+7)^(1/2) dx = (2x+7)^(1/2+1)/(1/2+1) + C
Simplifying, we get:
∫(2x+7)^(1/2) dx = (2x+7)^(3/2)/(3/2) + C
Now, we can solve for F(x) by substituting the original F(1) = 7 into the equation:
F(x) = ∫(2x+7)^(1/2) dx + C
F(x) = (2x+7)^(3/2)/(3/2) + C
Using F(1) = 7:
7 = (2(1)+7)^(3/2)/(3/2) + C
7 = (9)^(3/2)/(3/2) + C
To solve for the constant C, you need to simplify the fraction (9)^(3/2)/(3/2):
(9)^(3/2)/(3/2) = 9^(3/2) * 2/3 = 27 * 2/3 = 18
Substituting this value back into the equation:
7 = 18 + C
Solving for C:
C = 7 - 18 = -11
Therefore, the function F(x) becomes:
F(x) = (2x+7)^(3/2)/(3/2) - 11