Question

If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the absolute minimum and M is the absolute maximum of f on the interval [a, b], then m(b − a) ≤ b f(x) dx a ≤ M(b − a). Use this property to estimate the value of the integral. π/12 5 tan(4x) dx π/16

Answer 1

The correct question is:

If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the absolute minimum and M is the absolute maximum of f on the interval [a, b], then

m(b − a) ≤ (integral from a to b)f(x)dx ≤ M(b − a).

Use this property to estimate the value of the (integral from π/16 to π/12) 5 tan(4x)dx.

Answer:

The estimated value of the integral is 0.447030768

Step-by-step explanation:

Given f(x) = 5tan(4x)

f(x) is an increasing function on the interval (-infinity, infinity)

This implies that x must be in the interval (-π/8, π/8)............................(1)

But we are given the interval

[π/16, π/12] ....................................(2)

Since this interval is contained in (1), we have π/16 to be the minimum, and π/12 to be the maximum.

Now, the minimum value is

f(π/16) = 5tan(4π/16)

= 5tan(π/4)

= 5 × 1

m = 5

Maximum value is

f(π/12) = 5tan(4π/12)

= 5tan(π/3)

= 5 × √3

M = 5√3

Now, because

m(b - a) ≤ (integral from a to b) f(x)dx ≤ M(b - a)

We have

5(π/12 - π/16) ≤ (Integral from (π/16 to π/12) 5tan(4x)dx ≤ 5√3(π/12 - π/16)

5π/48 ≤ (Integral from (π/16 to π/12) 5tan(4x)dx ≤ 5√3π/48

Let's take the midpoint of this interval to be the approximate value of the integral.

(Integral from (π/16 to π/12) 5tan(4x)dx is approximately

(5π/48 + 5√3π/48)/2

= 0.447030768