Question
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
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
Answers
Oops. My bad.
I missed f(x).
You are correct.
I missed f(x).
You are correct.
correct
Related Questions
let R be the region bounded by the graphs of y = sin(pie times x) and y = x^3 - 4.
a) find the ar...
1. Let R be the region in the first quadrant enclosed by the graphs of y=4-X , y=3x , and the y-axi...
3). The shaded region is bounded by the y-axis and the graphs of y=1+√x, y=2. Find the volume of the...