Antiderivative Definite Integral And Average Value

Question

I. The antiderivative of the function $$4*x - 1$$ is $$2*x^2 - x$$.

II. Evaluating the definite integral $$\int_0^2 (4*x - 1)\,dx$$ gives $$6$$.

III. The average value of the function on the interval [0,2] is $$\frac{6}{2} = 3$$.

Which of the above statements is/are correct?
 
| Evaluation Step           | Calculation                                  |
|---------------------------|----------------------------------------------|
| Antiderivative            | $$2*x^2 - x$$                                |
| At $$x=2$$                | $$2*2^2 - 2 = 8 - 2 = 6$$                     |
| At $$x=0$$                | $$0$$                                       |
| Definite Integral Value   | $$6 - 0 = 6$$                                |
| Average Value             | $$\frac{6}{2} = 3$$                          |

Accepted Answer

I, II, and III

Answer

II and III only

Answer

I and II only

Rank	User	Correct Count	Attempt Count	Time	Score
#1	yeejayc	2	2	2m 12s	68
#2	daniel.rufener	2	2	2m 47s	33
#3	haneesh19.12	1	1	1m 43s	-3

Evaluation Step	Calculation
Antiderivative	$$2*x^2 - x$$
At $$x=2$$	$$2*2^2 - 2 = 8 - 2 = 6$$
At $$x=0$$	$$0$$
Definite Integral Value	$$6 - 0 = 6$$
Average Value	$$\frac{6}{2} = 3$$

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