Find the area under the curve below from x = 0 to x = 2. Give your answer correct to 3 decimal places.
y = 2x - x2
**i'll just retype it to make the integration part clearer:
integral 2x = [2x^(1+1)]/(1+1) = (2x^2)/2 = x^2
integral x^2 = [x^(2+1)]/(2+1) = (x^3)/3
To find the area under the curve of the function y = 2x - x^2 from x = 0 to x = 2, we can use integration.
First, let's rewrite the equation as y = -x^2 + 2x.
To find the area, we need to evaluate the definite integral of this function with respect to x from 0 to 2.
∫[-x^2 + 2x] dx
To find the integral, we can use the power rule, which states that for any term of the form x^n, the integral is given by (x^(n+1))/(n+1). So, let's integrate term by term:
∫-x^2 dx + ∫2x dx
Integrating each term, we get:
-(x^3)/3 + (x^2) + C
(where C is the constant of integration).
Now, we can evaluate this expression from x = 0 to x = 2:
[-(2^3)/3 + (2^2)] - [-(0^3)/3 + (0^2)]
Simplifying:
[-8/3 + 4] - [0/3 + 0]
[-8/3 + 12/3] - [0/3]
4/3
Therefore, the area under the curve from x = 0 to x = 2 is 4/3.
Rounded to 3 decimal places, the answer is approximately 1.333.
To find the area under the curve of a function, you can use calculus and integrate the function over the given interval. In this case, we need to find the area under the curve of the function y = 2x - x^2 from x = 0 to x = 2.
First, we need to find the antiderivative (also known as the indefinite integral) of the function y = 2x - x^2. The antiderivative of 2x is x^2, and the antiderivative of -x^2 is -x^3/3. So, the antiderivative of 2x - x^2 is x^2 - x^3/3.
To find the definite integral over the interval [0, 2], we subtract the value of the antiderivative at the upper limit (x = 2) from the value at the lower limit (x = 0):
∫[0, 2] (2x - x^2) dx = [x^2 - x^3/3] from 0 to 2
Plugging in the upper limit into the antiderivative expression:
= (2^2 - (2^3)/3) - (0^2 - (0^3)/3)
Simplifying further:
= (4 - 8/3)
= 12/3 - 8/3
= 4/3
Thus, the area under the curve of y = 2x - x^2 from x = 0 to x = 2 is 4/3. Rounded to 3 decimal places, the answer is approximately 1.333.
is y = 2x - x^2 the function given? if it is,
area under the curve means the integral at these particular bounds,, first you have to recall how to integrate:
for ax^n (a is constant), add 1 to the power, then this n+1 must be divided:
integral (ax^n) = (a)[x^(n+1)]/(n+1)
thus for 2x - x^2, we do it separately (by term):
integral 2x = 2x^(1+1)/(1+1) = x^2
integral x^2 = x^(2+1)/(2+1) = (x^3)/3
then evaluate x^2 - (x^3)/3 from 0 to 2.
hope this helps. :)