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irakobra [83]
1 year ago
12

Use left-endpoint approximation to estimate the area under the curve of f(x)=x^3 on [1,3]. Use n = 4. Using the same function us

e midpoint approximation to estimate the area.
Mathematics
1 answer:
natulia [17]1 year ago
6 0

Split up the interval [1, 3] into 4 intervals of equal length,

\Delta x = \dfrac{3 - 1}4 = \dfrac12

The left endpoint of the i-th interval is

\ell_i = 1 + \dfrac i2

The area under the curve is then approximately

\displaystyle \int_1^3 f(x) \, dx \approx \sum_{i=1}^4 f(\ell_i)\Delta x \\\\ ~~~~~~~~ = \frac12 \sum_{i=1}^4 \left(1 + \frac i2\right)^3 \\\\ ~~~~~~~~ = \frac12 \sum_{i=1}^4 \left(1 + \frac{3i}2 + \frac{3i^2}4 + \frac{i^3}8\right) = \boxed{27}

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3 years ago
Which one doesn’t belong? Why? Explain. please help me
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5 0
3 years ago
Using L’hopital’s rule please I need this ASAP Is due today!!!!!!!!!
Gelneren [198K]

Answer:

See below.

Step-by-step explanation:

You differentiate  top and bottom of the fraction until substitution gives you a value.

I can do the third one for you:

Lim x --> 0 of sin2x / sin3x

= lim x --> 0 of 2 cos2x / 3 cos 3x

= 2 cos 0 / 3 cos 0

= 2/3.

Limit as x--> 0  of (e^x - (1 - x) / x

= limit as x --> 0 of e^x + x - 1 / x

=  lim (e^x + 1) / 1

= 1 + 1 / 1

= 2.

limit as x-->  00  of  3x^2 - 2x +  1/ (2x^2 + 3)

= limit as x --> 00 of  6x - 2 / 4x     ( 00 = infinity)

Applying l'hopitals rule again:

limit is 6 / 4 =  3/2.

Limit  as x --> 00 of (ln x)^3 / x

= limit  3 (Ln x)^2 ) / x

= limit of  6 ln x / x

= limit  6 / x

= 0.

We had  to apply l'hopitals  rule 3 times here,

5 0
3 years ago
I need help with one 1+16=14+2 is this true or false
Vlad1618 [11]

Answer:

That my friend is FALSE

Step-by-step explanation:

To figure something like this out you just need to add the numbers on both sides. If they end up to be the same thing then it is true. If not, its false.

<u />

<u>1+16</u>=14+2

17=<u>14+2</u>

17=16

Nope. 17 does not equal 16. So it is false.

3 0
4 years ago
Read 2 more answers
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