Find the area of the region between the graphs of f(x)=3x+8 and
g(x)=x^2 + 2x+2 over [0,2].
I got 34/3.
Calculus - Steve
∫[0,2] (x^2+2x+2) dx
= 1/3 x^3 + x^2 + 2x [0,2]
= 8/3 + 4 + 4
= 32/3
Why are you taking the antiderivative of x^2 +2x+2 when we are trying to find the area between the two graphs? Am I supposed to to take the area of the top graph and subtract the area if the bottom graph to find the area between two curves at [0,2]?
Here is my work:
∫[0,2] 3x+8-(x^2+2x+2)dx=
∫[0,2] -x^2 +x+6 dx=
[1/3 x^3 +1/2 x^2+6x] from 0 to 2= 34/3
Oops. My bad.
I missed f(x).
You are correct.
correct
I apologize for the confusion. You are correct, we need to find the area between the two curves by taking the integral of the top function minus the integral of the bottom function over the given interval [0, 2]. Your approach is the correct one.
The area between two curves is given by the definite integral of the difference between the two functions over the interval of interest.
In this case, we have f(x) = 3x + 8 as the top function and g(x) = x^2 + 2x + 2 as the bottom function.
To find the area between the two curves over the interval [0, 2], we need to evaluate the integral of (3x + 8) - (x^2 + 2x + 2) with respect to x from 0 to 2:
∫[0,2] (3x + 8) - (x^2 + 2x + 2) dx
Simplifying the integrand:
∫[0,2] 3x + 8 - x^2 - 2x - 2 dx
Combine like terms:
∫[0,2] -x^2 + x + 6 dx
Now we can integrate term by term:
∫[0,2] -x^2 dx + ∫[0,2] x dx + ∫[0,2] 6 dx
Integral of -x^2 with respect to x is -(1/3)x^3:
= -(1/3)x^3 from 0 to 2 + (1/2)x^2 from 0 to 2 + 6x from 0 to 2
Evaluating the definite integrals at the upper and lower limits:
= (-(1/3)(2)^3 + (1/2)(2)^2 + 6(2)) - (-(1/3)(0)^3 + (1/2)(0)^2 + 6(0))
Simplifying:
= (-8/3 + 2 + 12) - (0 + 0 + 0)
= -8/3 + 2 + 12
= 34/3
So, the area between the two curves over the interval [0, 2] is indeed 34/3.