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The area under the curve $y = x$ from $x = 0$ to $x = 4$ is:
A$4$, equal to the length of the interval on the chart
B$2$, half the upper limit on the chart at the value $4$
C$8$, since $\int_0^4 x\, dx = x^2/2 |_0^4 = 8$
D$16$, ignoring the factor of $1/2$ in the antiderivative
Answer & Solution
Correct answer: C. $8$, since $\int_0^4 x\, dx = x^2/2 |_0^4 = 8$
$\int_0^4 x\, dx = [x^2/2]_0^4 = 16/2 = 8$.
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