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Prove that if the integral of nonnegative function is zero then the function is zero

by RoRi

Let f be an integrable, non-negative function on the interval [a,b]. Prove that if

    \[ \int_a^b f(x) \, dx = 0 \]

then f(x) = 0 at each point x \in [a,b] at which f is continuous.

Proof. The proof is by contradiction. Let p \in [a,b] be a point at which f is continuous, and suppose f(p) \neq 0. This implies f(p) > 0 since f is non-negative by hypothesis. Since f is continuous at p, we know by the sign-preserving property of continuous functions (Theorem 3.7 in Apostol), that there is some neighborhood about p, say (p - \delta, p + \delta) such that f(x) has the same sign as f(p) for all x \in (p - \delta, p + \delta), i.e., such that f(x) > 0 for all x \in (p - \delta, p + \delta). But then by the monotone property of the integral we know,

    \[ \int_{p - \delta}^{p + \delta} f(x) \, dx > \int_{p - \delta}^{p + \delta} 0 \,dx = 0.\]

But then,

    \[ \int_a^b f(x) \,dx = \int_a^{p - \delta} f(x)\, dx + \int_{p - \delta}^{p + \delta} f(x) \, dx + \int_{p + \delta}^b f(x) \, dx = 0. \]

This is a contradiction since f is nonnegative we know

    \[ \int_a^{p - \delta} f(x) \, dx \geq 0 \qquad \text{and} \qquad \int_{p + \delta}^b f(x) \, dx \geq 0. \]

Since \int_{p - \delta}^{p + \delta} f(x) \, dx is strictly positive, the sum of the three pieces of the integral \int_a^b f(x) \, dx cannot equal 0. \qquad \blacksquare

Use the weighted mean value theorem to establish an integral formula

by RoRi

Prove that for any positive integer n,

    \[ \int_{\sqrt{n \pi}}^{\sqrt{(n+1)\pi}} \sin (t^2) \, dt = \frac{(-1)^n}{c}, \qquad \text{for } \sqrt{n \pi} \leq c \leq \sqrt{(n+1)\pi}. \]

Recall the weighted mean value theorem:

For functions f and g continuous on [a,b], if g never changes sign in [a,b] then there exists c \in [a,b] such that

    \[ \int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx. \]

Proof. In order to apply the weighted mean value theorem (Theorem 3.16 of Apostol) we need to identify the functions f and g of the theorem. So, let

    \[ f(t) = \frac{1}{t}, \qquad g(t) = t \sin (t^2). \]

Then,

    \[ \int_{\sqrt{n \pi}}^{\sqrt{(n+1)\pi}} \sin (t^2) \, dt = \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} \frac{t \sin (t^2)}{t} \, dt = \int_{\sqrt{n \pi}}^{\sqrt{(n+1)\pi}} f(t) g(t) \, dt. \]

Furthermore, for a positive integer n, g(t) = t \sin (t^2) does not change sign on the interval [\sqrt{n \pi}, \sqrt{(n+1)\pi}] (since \sin(t^2) and t do not change sign on this interval). So we may apply the theorem to conclude there exists c \in [\sqrt{n \pi}, \sqrt{(n+1)\pi} ] such that

    \[ \int_{\sqrt{n \pi}}^{(n+1)\pi} f(t) g(t) \, dt = f(c) \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} g(t) \, dt = \frac{1}{c} \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} t \sin (t^2) \, dt. \]

Now we need to evaluate this integral. (Here I’m going to cheat a little. In the book we do not yet have techniques available to evaluate this integral. We’ll develop the Fundamental Theorem of Calculus, etc, in Chapter 5 that will allow us to do this. If you have a way to do this without directly evaluating an integral that we don’t yet know how to evaluate please do leave a comment with your solution.)

    \[ \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} t \sin(t^2) \, dt = -\frac{1}{2} \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} -2 t \sin(t^2) \, dt = \left. -\frac{1}{2} \cos(t^2) \right|_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} = (-1)^n. \]

Where the final equality follows since \cos \left( \sqrt{(n+1) \pi}\right) - \cos \left( \sqrt{n \pi} \right) = -2 if n is even and equals 2 if n is odd. Then multiplying that by the -\frac{1}{2}, we get (-1)^n.

Putting this together we then have,

    \[ \int_{\sqrt{n \pi}}^{\sqrt{(n+1) \pi}} \sin (t^2) \, dt = \frac{(-1)^n}{c} \qquad c \in \left[ \sqrt{n \pi}, \sqrt{(n+1) \pi} \right]. \qquad \blacksquare\]

Identify and explain why a statement is false

by RoRi

Given two statements:

  1. The integral

        \[ \int_{2 \pi}^{4 \pi} \frac{\sin t}{t} \, dt > 0 \]

    because

        \[ \int_{2 \pi}^{3 \pi} \frac{\sin t}{t} \, dt > \int_{3 \pi}^{4 \pi} \frac{|\sin t|}{t} \, dt. \]

  2. The integral

        \[ \int_{2 \pi}^{4 \pi} \frac{\sin t}{t} \, dt = 0 \]

    because, by the weighted mean value theorem (Theorem 3.16 in Apostol), there exists a c in [2 \pi, 4 \pi] such that

        \[ \int_{2 \pi}^{4 \pi} \frac{ \sin t}{t} \, dt = \frac{1}{c} \int_{2 \pi}^{4 \pi} \sin t \, dt = \frac{ \cos (2\pi) - \cos (4 \pi)}{c} = 0. \]

    3. Identify which of the two statements is false, and explain why.

Statement (b) is false since we may not apply the weighted mean value theorem in this case. The weighted mean value theorem requires the function g(t) to not change sign on the interval [2 \pi, 4 \pi]. Since g(t) = \sin t does change sign on the interval, we cannot apply the theorem.

Use the weighted mean value theorem to approximate an integral

by RoRi

Note that

    \[ 1+x^6 = (1+x^2)(1-x^2 + x^4), \]

and use the weighted mean value theorem (Theorem 3.16 in Apostol) to prove that for positive a:

    \[ \frac{1}{1+a^6} \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right) \leq \int_0^a \frac{dx}{1+x^2} \leq a - \frac{a^3}{3} + \frac{a^5}{5}. \]

Using a = \frac{1}{10} compute the integral to six decimal places.

Recall the weighted mean value theorem:

For functions f and g continuous on [a,b], if g never changes sign in [a,b] then there exists c \in [a,b] such that

    \[ \int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx. \]

Proof. Let

    \[ f(x) = \frac{1}{1+x^6}, \qquad g(x) = 1 - x^2 + x^4. \]

Then,

    \begin{align*} \int_0^a \frac{dx}{1+x^2} &= \int_0^a \frac{1-x^2+x^4}{1+x^6} \, dx \\ &= \int_0^a f(x) g(x) \, dx \\ &= \frac{1}{c} \int_0^a g(x) \, dx \end{align*}

for some c \in [0,a]. Since f is strictly decreasing on [0,a], we know f(a) \leq f(c) \leq f(0); thus,

    \[ \frac{1}{1+a^6} \leq f(c) \leq 1. \]

Furthermore,

    \[ \int_0^a g(x) \, dx = \int_0^a (1-x^2 + x^4) \, dx = a - \frac{a^3}{3} + \frac{a^5}{5}. \]

Thus,

    \[ \left( \frac{1}{1+a^6} \right) \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right) \leq \int_0^a \frac{dx}{1+x^2} \leq \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right). \qquad \blacksquare\]

Now, taking a = \frac{1}{10} we compute,

    \begin{align*} && (.9999999)(.0996687) \leq \int_0^a \frac{dx}{1+x^2} \leq .0996687 \\ \implies && .0996686 \leq \int_0^a \frac{dx}{1+x^2} \leq .0996687 \\ \implies && \int_0^a \frac{dx}{1+x^2} \approx .099669. \end{align*}

Use the weighted mean value theorem to establish some inequalities

by RoRi

Note that

    \[ \sqrt{1-x^2} = \frac{1-x^2}{\sqrt{1-x^2}}, \]

and use the weighted mean value theorem (Theorem 3.16 in Apostol) to prove:

    \[ \frac{11}{24} \leq \int_0^{\frac{1}{2}} \sqrt{1-x^2} \, dx \leq \frac{11}{24} \sqrt{\frac{4}{3}}. \]

Recall the weighted mean value theorem:

For functions f and g continuous on [a,b], if g never changes sign in [a,b] then there exists c \in [a,b] such that

    \[ \int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx. \]

Proof. Let

    \[ f(x) = \frac{1}{\sqrt{1-x^2}}, \qquad g(x) = 1-x^2. \]

Then,

    \begin{align*} \int_0^{\frac{1}{2}} \sqrt{1-x^2} \, dx &= \int_0^{\frac{1}{2}} \frac{1-x^2}{\sqrt{1-x^2}} \, dx \\ &= \int_0^{\frac{1}{2}} f(x) g(x) \, dx \\ &= f(c) \int_0^{\frac{1}{2}} g(x) \, dx \end{align*}

for some c \in \left[ 0, \frac{1}{2} \right]. Since f is strictly increasing on \left[ 0, \frac{1}{2} \right], we have f(0) \leq f(c) \leq f\left( \frac{1}{2} \right) implies 1 \leq f(c) \leq \sqrt{\frac{4}{3}}. Thus,

    \[ f(0) \int_0^{\frac{1}{2}} g(x) \, dx \leq f(c) \int_0^{\frac{1}{2}} g(x) \, dx \leq f(1) \int_0^{\frac{1}{2}} g(x) \, dx. \]

We also know

    \[ \int_0^{\frac{1}{2}} g(x) \, dx = \int_0^{\frac{1}{2}} (1-x^2) \, dx = \frac{1}{2} - \frac{1}{24} = \frac{11}{24}. \]

So, putting these together,

    \[ \frac{11}{24} \leq \int_0^{\frac{1}{2}} \sqrt{1-x^2} \, dx \leq \frac{11}{24} \sqrt{\frac{4}{3}}. \]

Use the weighted mean value theorem to establish an inequality

by RoRi

Use the weighted mean value theorem (Theorem 3.16 in Apostol) to prove:

    \[ \frac{1}{10 \sqrt{2}} \leq \int_0^1 \frac{x^9}{\sqrt{1+x}} \, dx \leq \frac{1}{10}. \]

Recall the weighted mean value theorem:

For functions f and g continuous on [a,b], if g never changes sign in [a,b] then there exists c \in [a,b] such that

    \[ \int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx. \]

Proof. Let

    \[ f(x) = \frac{1}{\sqrt{1+x}}, \qquad g(x) = x^9. \]

Then substituting our definitions of f and g,

    \[ \int_0^1 \frac{x^9}{\sqrt{1+x}} \, dx = \int_0^1 f(x) g(x) \, dx. \]

Since f and g are continuous and g does not change sign on [0,1] we may apply Theorem 3.16,

    \[ \int_0^1 f(x) g(x) \, dx = f(c) \int_0^1 g(x) \, dx \qquad \text{for some } c \in [0,1]. \]

Since f is strictly decreasing on [0,1], we have

    \[ f(0) \geq f(c) \geq f(1) \quad \implies \quad 1 \geq f(c) \geq \frac{1}{\sqrt{2}}. \]

Then,

    \begin{align*} &&f(1) \int_0^1 g(x) \, dx &\leq f(c) \int_0^1 g(x) \, dx \leq f(0) \int_0^1 g(x) \, dx \\[8pt] \implies && \frac{1}{\sqrt{2}} \int_0^1 x^9 \, dx &\leq f(c) \int_0^1 x^9 \, dx \leq \int_0^1 x^9 \, dx \\[8pt] \implies && \frac{1}{10 \sqrt{2}} &\leq f(c) \int_0^1 \frac{x^9}{\sqrt{1+x}} \, dx \leq \frac{1}{10}. \qquad \blacksquare \end{align*}