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Chapter 2: Problem 25

Let \(\mathrm{f}\) be twice continuously differentiable in \([0,2 \pi]\) and concave up. Prove that \(\int_{0}^{2 \pi} f(x) \cos x d x \geq 0\)

Short Answer

Expert verified
Question: Prove that for a twice continuously differentiable function f that is concave up in the interval [0, 2蟺], the following inequality holds: 鈭個虏蟺 f(x) cos x dx 鈮� 0. Answer: By using integration by parts and the property of concave up functions, we showed that the given integral is equal to the integral of the product of the second derivative of f and the cosine function. Since the function is concave up, its second derivative is positive, and given the properties of the cosine function, the integral is non-negative. Therefore, 鈭個虏蟺 f(x) cos x dx 鈮� 0.

Step by step solution

01

Recall the definition of concave up

A function is concave up in a certain interval if its second derivative is positive in that interval. In our case, the function \(\mathrm{f}\) is concave up in \([0, 2\pi]\), which implies that \(f''(x) > 0\) for all \(x \in [0, 2\pi]\).
02

Use integration by parts to evaluate the integral of f(x) cos(x)

Integration by parts states that \(\int u dv = uv - \int v du\), where \(u\) and \(v\) are differentiable functions in the interval in question. For our case, let's choose \(u=f(x)\) and \(dv = \cos x dx\). Applying differentiation and integration to obtain 'du' and 'v', let: \(du = f'(x)dx\) \(v = \int \cos x dx = \sin x\) Now using integration by parts formula: \(\int_{0}^{2 \pi} f(x) \cos x dx = f(x) \sin x \Big|_0^{2 \pi} - \int_{0}^{2 \pi} f'(x) \sin x dx\) Notice that for any twice differentiable function f(x), \(f(x) \sin x \Big|_0^{2 \pi} = f(2\pi)\sin(2\pi) - f(0)\sin(0) = 0\).
03

Apply integration by parts again

Now, we will apply integration by parts again to the integral that remains. Let's choose \(u=f'(x)\) and \(dv = \sin x dx\). Then, we have: \(du = f''(x)dx\) \(v = \int \sin x dx = -\cos x\) So, the remaining integral becomes: \(-\int_{0}^{2 \pi} f'(x) \sin x dx = -f'(x)(-\cos x) \Big|_0^{2 \pi} - \int_{0}^{2 \pi} f''(x)(-\cos x)dx\) Again, we have, \(f'(2 \pi)(-\cos 2\pi) - f'(0)(-\cos 0) = 0\). Therefore, our final result is: \(\int_{0}^{2 \pi} f(x) \cos x dx = 0 + 0 - \int_{0}^{2 \pi} f''(x)(-\cos x)dx = \int_{0}^{2 \pi} f''(x) \cos x dx\).
04

Show that the integral is non-negative

Since the function \(\mathrm{f}\) is concave up in \([0,2\pi]\), we know that its second derivative is positive in that interval (\(f''(x) > 0\) for all \(x \in [0,2\pi]\)). Moreover, \(\cos x \geq 0\) for all \(x \in [0, \pi]\) and \(\cos x \leq 0\) for all \(x \in [\pi, 2\pi]\). Hence, \(\int_{0}^{2 \pi} f''(x) \cos x dx = \int_{0}^{\pi} f''(x) \cos x dx + \int_{\pi}^{2\pi} f''(x) \cos x dx \geq 0\). We have achieved our goal and shown that: \(\int_{0}^{2 \pi} f(x) \cos x dx \geq 0\).

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