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Dmitry_Shevchenko [17]
3 years ago
12

Find the approximate area between the curve f(x) = -4x² + 32x and on the x-axis on the interval [0,8] using 4 rectangles. Use th

e right endpoint of each rectangle to determine the height.

Mathematics
1 answer:
Doss [256]3 years ago
6 0

Split up the interval [0, 8] into 4 equally spaced subintervals:

[0, 2], [2, 4], [4, 6], [6, 8]

Take the right endpoints, which form the arithmetic sequence

r_i=2+\dfrac{8-0}4(i-1)=2i

where 1 ≤ <em>i</em> ≤ 4.

Find the values of the function at these endpoints:

f(r_i)=-4{r_i}^2+32r_i=-16i^2+64i

The area is given approximately by the Riemann sum,

\displaystyle\int_0^8f(x)\,\mathrm dx\approx\sum_{i=1}^4f(r_i)\Delta x_i

where \Delta x_i=\frac{8-0}4=2; so the area is approximately

\displaystyle2\sum_{i=1}^4(-16i^2+64i)=-32\sum_{i=1}^4i^2+128\sum_{i=1}^4i=-32\cdot\frac{4\cdot5\cdot9}6+128\cdot\frac{4\cdot5}2=\boxed{320}

where we use the formulas,

\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2

\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6

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Option 3

y + 16 = 8(x + 2)

Step-by-step explanation:

<h3><u>Given</u>;</h3>
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  • y = 16, 32, 48, 64

Now, Put the values in equation 1

y – 16 = 8(x – 2)

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y – 16 = 8(x – 2)

16 – 16 = 8(2 – 2)

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Here, L.H.S ≠ R.H.S

So, Option 1 is incorrect

Now, Put the values in equation 2

y – 16 = 8x – 2

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y – 16 = 8x – 2

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Here, L.H.S ≠ R.H.S

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y + 16 = 8(x + 2)

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y + 16 = 8(x + 2)

16 + 16 = 8(2 + 2)

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Here, L.H.S = R.H.S

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y + 16 = 8(x + 2)

32 + 16 = 8(4 + 2)

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48 = 48

Here, L.H.S = R.H.S

  • For x = 6 and y = 48

y + 16 = 8(x + 2)

48 + 16 = 8(6 + 2)

64 = 8(8)

64 = 64

Here, L.H.S = R.H.S

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y + 16 = 8(x + 2)

64 + 16 = 8(8 + 2)

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80 = 80

Here, L.H.S = R.H.S

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