$\lim_{n \to \infty} \sum\limits_{k=1}^{n}f(x_{k}) \Delta x = \int\limits^a_b {f(x)} \, dx \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times k$

In this case we have:

Δx = 3/n

b − a = 3

a = 1

b = 4

So the integral is:

∫₁⁴ √x dx

To evaluate the integral, we write the radical as an exponent.

∫₁⁴ x^½ dx

= ⅔ x^³/₂ + C |₁⁴

= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)

= ⅔ (8) + C − ⅔ − C

= 14/3

If ∫₁⁴ f(x) dx = e⁴ − e, then:

∫₁⁴ (2f(x) − 1) dx

= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx

= 2 (e⁴ − e) − (x + C) |₁⁴

= 2e⁴ − 2e − 3

∫ sec²(x/k) dx

k ∫ 1/k sec²(x/k) dx

k tan(x/k) + C

Evaluating between x=0 and x=π/2:

k tan(π/(2k)) + C − (k tan(0) + C)

k tan(π/(2k))

Setting this equal to k:

k tan(π/(2k)) = k

tan(π/(2k)) = 1

π/(2k) = π/4

1/(2k) = 1/4

2k = 4

k = 2

8 0

3 years ago

Mai needs to make a total of 80 deliveries this week. So far she has completed 60 of them. What percentage of her total deliveri

prisoha [69]

Answer:

THX FOR POINTS

Step-by-step explanation:

4 0

2 years ago