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.
The integral from 0 to 2 of 1/1+x^2
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To estimate the value of the integral ∫(0 to 2) 1/(1+x²), we need to use the property given to us.
First, we need to find the absolute minimum and maximum of f(x) on the interval [0, 2]. Let's find the derivatives of f(x) and check for critical points:
f(x) = 1/(1+x²)
f'(x) = (-2x)/(1+x²)²
To find critical points, we set f'(x) = 0 and solve for x:
(-2x)/(1+x²)² = 0
Since the numerator is 0, we have -2x = 0, which implies x = 0.
Now, let's check the endpoints of the interval [0, 2]:
f(0) = 1/(1+(0)²) = 1/1 = 1
f(2) = 1/(1+(2)²) = 1/5
Therefore, the absolute minimum m = f(2) = 1/5 and the absolute maximum M = f(0) = 1.
Now, we can use the property m(b-a) ≤ ∫(a to b) f(x) dx ≤ M(b-a) to estimate the value of the integral:
m(b-a) = (1/5)(2-0) = 2/5
M(b-a) = (1)(2-0) = 2
Therefore, the estimated value of the integral ∫(0 to 2) 1/(1+x²) is between 2/5 and 2.