Exam scores in a large class are normally distributed with a mean of 60
1 answer:
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Step-by-step explanation:
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
