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Wittaler [7]
2 years ago
15

HELPP!!! 100 POINTS AND BRAINLIEST!! Sketch the graph of the function y = 20x − x2, and approximate the area under the curve in

the interval [0, 20] by dividing the area into the given numbers of rectangles.
Part A
Use five rectangles to approximate the area under the curve.
Part B
Use 10 rectangles to approximate the area under the curve.
Part C
Calculate the area under the curve using rectangles as their number becomes arbitrarily large (tends to infinity). You do not need to sketch the rectangles.
Mathematics
1 answer:
Arte-miy333 [17]2 years ago
5 0

Answer:

Hello,

Step-by-step explanation:

Part A: see picture

Part B: see picture

Part C:

You must remenber :

\displaystyle \sum_{i=1}^{n} i= \dfrac{n*(n+1)}{2} \\\\\displaystyle \sum_{i=1}^{n} i^2= \dfrac{n*(n+1)(2n+1)}{6} \\\\\\\int\limits^{20}_0 {(20\ x-x^2)} \, dx =[\dfrac{20\ x^2}{2} -\dfrac{x^3}{3} ]^{20}_0=\dfrac{20^3}{3} =1333.\overline{3}\\\\\\

We divide the intervalle in n equal parts:

f(x)=20\ x- x^2\\\Delta x=\dfrac{20-0}{n} =\dfrac{20}{n} \\Area\ of\ a\ rectangle\ R_i=\Delta x*f(x_i)\\x_i=0+\Delta x*i=\Delta x*i\\\\\\\displaystyle \int\limits^{20}_0 {(20 \ x-x^2)} \, dx = \lim_{n \to \infty}  \sum_{i=1}^n(\Delta x* f(x_i) )\\\\=\lim_{n \to \infty}  \sum_{i=1}^n(\Delta x* (20*\Delta x*i-(\Delta x*i)^2)\\\\=\lim_{n \to \infty} (20*(\dfrac{20}{n})^2\sum_{i=1}^n(i)-(\dfrac{20}{n})^3\sum_{i=1}^n(i^2)\\\\

\displaystyle =\lim_{n \to \infty}(20*(\dfrac{20}{n})^2*\dfrac{n*(n+1)}{2} -(\dfrac{20}{n})^3*\dfrac{n*(n+1)(2n+1)}{6} \\\\=10*20^2-\dfrac{20^3}{6} *2\\\\\\=4000-\frac{8000}{3} \\\\=\dfrac{4000}{3} \\\\=1333,\overline{3}\\\\

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How do you do number 1?<br>Whenever I tried to answer it, I always get fraction. help me.​
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Answer:

The pairs are (13,15) and (-15,-13).

Step-by-step explanation:

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n+1 is even (so we aren't using this number)

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This means

(n)^2+(n+2)^2=394

n^2+(n+2)(n+2)=394

n^2+n^2+4n+4=394               since (a+b)(a+b)=a^2+2ab+b^2

Combine like terms:

2n^2+4n+4=394

Subtract 394 on both sides:

2n^2+4n-390=0

Divide both sides by 2:

n^2+2n-195=0

Now we need to find two numbers that multiply to be -195 and add up to be 2.

15 and -13 since 15(-13)=-195 and 15+(-13)=2

So the factored form is

(n+15)(n-13)=0

This means we have n+15=0 and n-13=0 to solve.

n+15=0

Subtract 15 on both sides:

n=-15

n-13=0

Add 13 on both sides:

n=13

So if n=13 , then n+2=15.

If n=-15, then n+2=-13.

Let's check both results

(n,n+2)=(13,15)

13^2+15^2=169+225=394.  So (13,15) looks good!

(n,n+2)=(-15,-13)

(-15)^2+(-13)^2=225+169=394.  So (-15,-13) looks good!

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