Integral from 0 to pi/6 of (sin x + cos x) less than or equal to pi(square root 3 + 1)/12
The integral is (-cos x + sin x)[0,π/6]
= (-√3/2 + 1/2) - (-1 + 0) = 3/2 - √3/2 = 0.63397
π(√3 + 1)/12 = 0.71525
the statement is true
It's kind of a weird question, since the integral value has nothing to do with π.
To solve the given integral inequality, we need to find the integral of the function "sin(x) + cos(x)" over the interval [0, π/6] and compare it to the value on the right side of the inequality, which is (π(√3 + 1))/12.
First, let's find the integral of "sin(x) + cos(x)" over the given interval. We can do this by taking the antiderivative of each term separately and applying the limits:
∫[0, π/6] (sin(x) + cos(x)) dx
= [-cos(x) + sin(x)] from 0 to π/6
= (-cos(π/6) + sin(π/6)) - (-cos(0) + sin(0))
= (-√3/2 + 1/2) - (-1 + 0)
= -√3/2 + 1/2 + 1
= -√3/2 + 3/2
Next, we compare this value to the right side of the inequality:
-√3/2 + 3/2 ≤ (π(√3 + 1))/12
To simplify the comparison, we can multiply both sides of the inequality by 12 to get rid of the denominator:
12 * (-√3/2 + 3/2) ≤ π(√3 + 1)
-6√3 + 18 ≤ π(√3 + 1)
Now, we isolate π on one side of the inequality by subtracting -6√3 + 18 from both sides:
-6√3 ≤ π(√3 + 1) - 18
Now, we have an upper bound for the value of π. By evaluating the right side of the inequality and dividing both sides by (√3 + 1), we can find a numerical approximation for π.
-6√3 ≤ π(√3 + 1) - 18
-6√3 + 18 ≤ π(√3 + 1)
(-6√3 + 18)/(√3 + 1) ≤ π
Using a calculator, we evaluate the expression on the left side and find:
π ≥ 3.23
Therefore, the inequality holds, and the original integral is less than or equal to (π(√3 + 1))/12.