After you made your sketch, you should realize that the curve is below the x-axis from 0 to 1 and above the x-axis from 1 to 3.
You will have to calculate it in two parts
Area = [integral](-x^2 + x) from 0 to 1 + [integral](x^2 - x) from 1 to 3
The rest is easy, take it from here
Find the area between the x-axis and the graph y = x^2 – x + 1 from x = 0 to x = 3.
3 answers
I don't get it..Do I need to plug in anything for x or no?
OOPS, I missed the +1 at the end of your function
Now your question becomes even easier. The curve is entirely above the x-axis
Area = integral [x^2 - x + 1] from 0 to 3
= [x^3/3 - x^2/2 + x] from 0 to 3
= 9 - 9/2 + 3 - 0
= 15/2 or 7.5
Now your question becomes even easier. The curve is entirely above the x-axis
Area = integral [x^2 - x + 1] from 0 to 3
= [x^3/3 - x^2/2 + x] from 0 to 3
= 9 - 9/2 + 3 - 0
= 15/2 or 7.5
0 answers