1. Area of the region between the graph of y=2x^2-3x and the x -axis from x=-3 to x=1 is
a)62/3
b)97/3
c)16/3
d)20/3
e)92/3
I got e but it's incorrect
I also got (e).
However, that is an algebraic area, which includes the sign.
If you want the geometric area, you have to split into two parts, because it crosses the x-axis at x=0.
Then you get 97/3
I agree with Steve, it is 97/3
Thank you!
To find the area between the graph of the given function and the x-axis in the given interval, you can use the definite integral. The area is given by:
A = ∫(b, a) [f(x)dx]
In this case, f(x) = 2x^2 - 3x is the function representing the graph, and the interval is from x = -3 to x = 1. So, we need to evaluate the integral:
A = ∫(1, -3) [2x^2 - 3x dx]
To calculate this definite integral, we can use the Fundamental Theorem of Calculus or integrate the function term by term:
A = [2/3 x^3 - 3/2 x^2] from x = -3 to x = 1
Substituting the upper and lower limits, we get:
A = [2/3 (1)^3 - 3/2 (1)^2] - [2/3 (-3)^3 - 3/2 (-3)^2]
= [2/3 - 3/2] - [-18/3 - 27/2]
= [2/3 - 9/6] - [-6 - 27/2]
= [4/6 - 9/6] - [-12 - 27/2]
= (-5/6) - (-12 - 27/2)
= -5/6 + 12 + 27/2
= -5/6 + 12 + 27/2
By finding the common denominator of 6 to add the fractions, we get:
A = -5/6 + 72/6 + 81/6
= 148/6
= 74/3
Therefore, the correct answer is not e) 92/3 but rather:
a) 62/3