select the correct expressions. identify each expression and value that represents the area under the curve…

select the correct expressions. identify each expression and value that represents the area under the curve y = -0.01x² + 40 on the interval -10, 10 396.67 square units limn→∞∑k = 1n{-0.01(-10 + 10k/n)² + 40}(10/n) square units limn→∞∑k = 1n(780/n + 80k/n² - 80k²/n³) square units limn→∞∑k = 1n(390/n + 20k/n² - 10k²/n³) square units limn→∞∑k = 1n-(780/n + 80k/n² - 80k²/n³) square units 846.67 square units 793.33 square units limn→∞∑k = 1n{-0.01(-10 + 20k/n)² + 40}(20/n) square units

select the correct expressions. identify each expression and value that represents the area under the curve y = -0.01x² + 40 on the interval -10, 10 396.67 square units limn→∞∑k = 1n{-0.01(-10 + 10k/n)² + 40}(10/n) square units limn→∞∑k = 1n(780/n + 80k/n² - 80k²/n³) square units limn→∞∑k = 1n(390/n + 20k/n² - 10k²/n³) square units limn→∞∑k = 1n-(780/n + 80k/n² - 80k²/n³) square units 846.67 square units 793.33 square units limn→∞∑k = 1n{-0.01(-10 + 20k/n)² + 40}(20/n) square units

Answer

Explanation:

Step1: Recall the definite - integral for area

The area (A) under the curve (y = f(x)) on the interval ([a,b]) is given by (A=\int_{a}^{b}f(x)dx), where (f(x)=- 0.01x^{2}+40), (a = - 10), and (b = 10). Also, the definite - integral (\int_{a}^{b}f(x)dx=\lim_{n\rightarrow\infty}\sum_{k = 1}^{n}f(x_{k})\Delta x), with (\Delta x=\frac{b - a}{n}) and (x_{k}=a + k\Delta x). Here, (a=-10), (b = 10), so (\Delta x=\frac{10-(-10)}{n}=\frac{20}{n}) and (x_{k}=-10+\frac{20k}{n}). Then (f(x_{k})=-0.01(-10+\frac{20k}{n})^{2}+40). And the area (A=\lim_{n\rightarrow\infty}\sum_{k = 1}^{n}\left[-0.01(-10+\frac{20k}{n})^{2}+40\right]\frac{20}{n}).

Step2: Expand (f(x_{k}))

First, expand (-0.01(-10+\frac{20k}{n})^{2}=-0.01(100 - \frac{400k}{n}+\frac{400k^{2}}{n^{2}})=-1+\frac{4k}{n}-\frac{4k^{2}}{n^{2}}). Then (f(x_{k})=-1+\frac{4k}{n}-\frac{4k^{2}}{n^{2}} + 40=39+\frac{4k}{n}-\frac{4k^{2}}{n^{2}}). And (\sum_{k = 1}^{n}f(x_{k})\Delta x=\sum_{k = 1}^{n}(39+\frac{4k}{n}-\frac{4k^{2}}{n^{2}})\frac{20}{n}=\sum_{k = 1}^{n}(\frac{780}{n}+\frac{80k}{n^{2}}-\frac{80k^{2}}{n^{3}})).

Step3: Calculate the definite - integral

(\int_{-10}^{10}(-0.01x^{2}+40)dx=\left[-0.01\times\frac{x^{3}}{3}+40x\right]_{-10}^{10}) [ \begin{align*} &(-0.01\times\frac{10^{3}}{3}+40\times10)-(-0.01\times\frac{(-10)^{3}}{3}+40\times(-10))\ =&(-\frac{10}{3}+400)-(\frac{10}{3}-400)\ =&-\frac{10}{3}+400-\frac{10}{3}+400\ =&800-\frac{20}{3}\ =&\frac{2400 - 20}{3}=\frac{2380}{3}\approx793.33 \end{align*} ]

Answer:

(\lim_{n\rightarrow\infty}\sum_{k = 1}^{n}\left[-0.01(-10+\frac{20k}{n})^{2}+40\right]\frac{20}{n}) square units, (\lim_{n\rightarrow\infty}\sum_{k = 1}^{n}(\frac{780}{n}+\frac{80k}{n^{2}}-\frac{80k^{2}}{n^{3}})) square units, (793.33) square units