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Maurinko [17]
3 years ago
8

Consider the integral 8 (x2+1) dx 0 (a) Estimate the area under the curve using a left-hand sum with n = 4. 250 Is this sum an o

verestimate or an underestimate of the true value? overestimate underestimate (b) Estimate the area under the curve using a right-hand sum with n = 4. 248

Mathematics
1 answer:
Leya [2.2K]3 years ago
3 0

Answer:

  (a) 120 square units (underestimate)

  (b) 248 square units

Step-by-step explanation:

<u>(a) left sum</u>

See the attachment for a diagram of the areas being summed (in orange). This is the sum of the first 4 table values for f(x), each multiplied by 2 (the width of the rectangle). Quite clearly, the curve is above the rectangle for the entire interval, so the rectangle area underestimates the area under the curve.

  left sum = 2(1 + 5 + 17 + 37) = 2(60) = 120 . . . . square units

<u>(b) right sum</u>

The right sum is the sum of the last 4 table values for f(x), each multiplied by 2 (the width of the rectangle). This sum is ...

  right sum = 2(5 +17 + 37 +65) = 2(124) = 248 . . . . square units

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Given:

$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}

To find:

The simplified fraction.

Solution:

Step 1: Simplify the numerator

$\frac{(4 r)^{3}}{15 t^{4}}=\frac{4^3 r^{3}}{15 t^{4}}=\frac{64 r^{3}}{15 t^{4}}

Step 2: Simplify the denominator

$\frac{16 r}{(3 t)^{2}}=\frac{16 r}{3^2 t^{2}}= \frac{16 r}{9 t^{2}}

Step 3: Using step 1 and step 2

$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}} \right)}

Step 4: Using fraction rule:

$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a \cdot d}{b \cdot c}

$\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}}\right)}=\frac{64r^3 \cdot 9t^2}{16 r \cdot 15 t^4}

Cancel the common factor r and t², we get

           $=\frac{64 r^{2} \cdot 9 }{16  \cdot 15 t^2 }

Cancel the common factors 16 and 3 on both numerator and denominator.

           $=\frac{4 r^{2} \cdot 3 }{  5 t^2 }

           $=\frac{12 r^{2}  }{  5 t^2 }

$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{12 r^{2}  }{  5 t^2 }

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5 0
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mafiozo [28]
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2 years ago
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Alexxandr [17]

Answer:

The area of the figure is equal to 35\ cm^{2}

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

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<u>Find the area of the square</u>

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<u>Find the area of the triangle</u>

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Answer:

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Step-by-step explanation:

That's Correct!

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