Find a one unit interval with the greatest area?
Consider the curve g(x) = sin(x²) for x in the interval (0,√pi). Use calculus (follow guideline below) to find a one-unit interval with the greatest area between the curve and the x-axis. Note: Do not do this problem by trial and error.
Use the following guideline with proper mathematical notation:
a) Set up a function for this area. This function may involve integrals.
b) Use calculus to maximize this function. Answer the question by giving your one unit interval. round answers to four decimal places.
Consider the curve g(x) = sin(x²) for x in the interval (0,√pi). Use calculus (follow guideline below) to find a one-unit interval with the greatest area between the curve and the x-axis. Note: Do not do this problem by trial and error.
Use the following guideline with proper mathematical notation:
a) Set up a function for this area. This function may involve integrals.
b) Use calculus to maximize this function. Answer the question by giving your one unit interval. round answers to four decimal places.
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Let the interval be (t, t+1)
Then the area under the curve is
A(t) = ∫sin(x²) dx from t to t+1.
Now I've never before done "differentiation under the integral sign" with both boundaries variable, so let's write this as
A(t) = ∫sin(x²) dx [t to 0] + ∫sin(x²) dx [0 to t+1]
Now reverse the boundaries for the first one, and for the second one, substitute x = u+1:
A(t) = -∫sin(x²) dx [0 to t] + ∫sin((u+1)²) du [u = -1 to t]
Now differentiation under the integral sign gives
A'(t) = - sin(t²) + sin((t+1)²)
Is there a solution to sin(t²) = sin((t+1)²) in the interval t =0 to t = √π?
These two sine functions are equal when
π - (t+1)² = t²
when 0 = 2t² + 2t - (π-1)
t = [-2 ±√(4 + 8(π-1))]/4
. ≈ 0.649259 [the other root is negative]
Since A'(0) > 0 and A'(√π) < 0,
the stationary point gives a maximum value.
Hence the required interval, correct to four decimal places, is
(0.6493, 1.6493)
Then the area under the curve is
A(t) = ∫sin(x²) dx from t to t+1.
Now I've never before done "differentiation under the integral sign" with both boundaries variable, so let's write this as
A(t) = ∫sin(x²) dx [t to 0] + ∫sin(x²) dx [0 to t+1]
Now reverse the boundaries for the first one, and for the second one, substitute x = u+1:
A(t) = -∫sin(x²) dx [0 to t] + ∫sin((u+1)²) du [u = -1 to t]
Now differentiation under the integral sign gives
A'(t) = - sin(t²) + sin((t+1)²)
Is there a solution to sin(t²) = sin((t+1)²) in the interval t =0 to t = √π?
These two sine functions are equal when
π - (t+1)² = t²
when 0 = 2t² + 2t - (π-1)
t = [-2 ±√(4 + 8(π-1))]/4
. ≈ 0.649259 [the other root is negative]
Since A'(0) > 0 and A'(√π) < 0,
the stationary point gives a maximum value.
Hence the required interval, correct to four decimal places, is
(0.6493, 1.6493)