Step-by-step explanation:
[tex]\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[/tex]
In this case we have:
Ξx = 3/n
b β a = 3
a = 1
b = 4
So the integral is:
β«ββ΄ βx dx
To evaluate the integral, we write the radical as an exponent.
β«ββ΄ x^Β½ dx
= β
x^Β³/β + C |ββ΄
= (β
4^Β³/β + C) β (β
1^Β³/β + C)
= β
(8) + C β β
β C
= 14/3
If β«ββ΄ f(x) dx = eβ΄ β e, then:
β«ββ΄ (2f(x) β 1) dx
= 2 β«ββ΄ f(x) dx β β«ββ΄ dx
= 2 (eβ΄ β e) β (x + C) |ββ΄
= 2eβ΄ β 2e β 3
β« secΒ²(x/k) dx
k β« 1/k secΒ²(x/k) dx
k tan(x/k) + C
Evaluating between x=0 and x=Ο/2:
k tan(Ο/(2k)) + C β (k tan(0) + C)
k tan(Ο/(2k))
Setting this equal to k:
k tan(Ο/(2k)) = k
tan(Ο/(2k)) = 1
Ο/(2k) = Ο/4
1/(2k) = 1/4
2k = 4
k = 2
