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jek_recluse [69]
3 years ago
9

The right-hand endpoint of the kth subinterval is denoted x∗k. What is x∗k (in terms of k and n)? x∗k =_________.

Mathematics
1 answer:
Strike441 [17]3 years ago
3 0

In this problem you will calculate ∫302+4 by using the formal definition of the definite integral: ∫()=lim→∞[∑=1(∗)Δ].

(a) The interval [0,3] is divided into equal subintervals of length Δ. What is Δ (in terms of )? Δ =

(b) The right-hand endpoint of the th subinterval is denoted ∗. What is ∗ (in terms of and )? ∗ =

Answer:

a) Δ= \frac{3}{n}

b) x^{*}_{k} = \frac{3k}{n}

Step-by-step explanation:

a)  If the interval [0,3] , i.e let a = 0 , b =3 and n=n.

So [0,3] divide into n equal subintervals;

Therefore, the length Δ= \frac{b-a}{n}

Δ= \frac{3-0}{n}

Δ= \frac{3}{n}

b) To calculate x^{*}_{k};

x^{*}_{k} = a + k . Δ          (where n= 0, Δ = \frac{3}{n})

= 0 + k . \frac{3}{n}

x^{*}_{k}  =  \frac{3}{k}

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3 years ago
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Hem dibuixat tres rectangles. En el primer, la llargada mesura 3 cm més que l'amplada. El segon i tercer rectangle tenen unes di
quester [9]

Resposta:

Primer rectangle:

Amplada = 11

Longitud = 14

Segon rectangle:

Amplada = 12

Longitud = 15

Tercer rectangle:

Amplada = 13

Longitud = 16

Explicació pas a pas:

Donat que:

Primer rectangle:

Amplada = x

Longitud = x + 3

2n rectangle:

Augment de la dimensió d'1 cm respecte al primer rectangle;

Amplada = x + 1

Longitud = x + 4

3r rectangle:

Augment de la dimensió de 2 cm respecte al primer rectangle;

Amplada = x + 2

Longitud = x + 5

Suma dels tres perímetres del rectangle:

Perímetre d'un rectangle: 2 (l + O)

Primer rectangle:

2 (x + x + 3) = 2 (2x + 3) = 4x + 6

2n:

2 (x + 1 + x + 4) = 2 (2x + 5) = 4x + 10

3r:

2 (x + 2 + x + 5) = 2 (2x + 7) = 4x + 14

Suma de perímetres = 162

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Per tant,

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2n rectangle:

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3r rectangle:

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The input of 8 is repeating.

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Answer:

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