Riemann Integral Exercises for undergraduate students

In this article, we offer a selection of Riemann integral exercises for undergraduate students. This is the first step for the more general Lebesgue integrals studied in mathematics.

Riemann integral exercises

Exercise: Let $f:[0,1]\to \mathbb{R}$ be a continuous function. For any $n\in\mathbb{N},$ we set \begin{align*} u_n=n\int^1_0 t^n f(t)dt. \end{align*}

  • Let $\varepsilon>0$. Prove that there exists $a\in (0,1)$ such that $|f(t)|\le \frac{\varepsilon}{2}$ whenever $t\in [a,1]$.
  • We denote $M=\sup\{|f(t)|:t\in [0,1]\}$ (a such $M\in \mathbb{R}^+$ exists because $f$ is continuous on the compact $[0,1]$). Prove that \begin{align*} |u_n|\le M a^n+\frac{\varepsilon}{2},\qquad \forall n\in\mathbb{N}. \end{align*}
  • Deduce that $\lim_{n\to+\infty}u_n=0$.
  • Deduce that, in general, we have \begin{align*} \lim_{n\to+\infty}u_n=f(1). \end{align*}

Solution: The first 3 question: By definition of left continuity at point $1,$ for all $\varepsilon>0,$ there exists $0 < \delta < 1$ such that $t\in [1-\delta,1]$ we have $|f(t)-f(1)|=|f(t)| \le \frac{\varepsilon}{2}$. It suffices to put $a=1-\delta\in (0,1)$.

>We can write \begin{align*} |u_n|&\le n \int^1_0 t^n |f(t)|dt\cr &\le n \int^a_0 t^n |f(t)|dt+n \int^1_a t^n |f(t)|dt\cr & \le n M \int^a_0 t^n dt+ n \frac{\varepsilon}{2} \int^1_a t^n dt\cr & \le n M \left[\frac{t^{n+1}}{n+1}\right]^a_0+ n \frac{\varepsilon}{2} \left[\frac{t^{n+1}}{n+1}\right]^1_a \cr &\le M \frac{n}{n+1} a^{n+1}+\frac{\varepsilon}{2} \frac{n}{n+1} (1-a^{n+1}). \end{align*} As $a\in (0,1),$ then $a^{n+1}\le a^n$ and $0 < 1-a^{n+1} < 1$. On the other hand, $\frac{n}{n+1}\le 1$. This implies that \begin{align*} |u_n|\le M a^n+\frac{\varepsilon}{2},\qquad \forall n\in\mathbb{N}. \end{align*}

Since $a\in (0,1)$ then the geometric sequence $(a^n)_n$ satisfies $a^n\to 0$ as $n\to+\infty$. Then there exists $N\in \mathbb{N}$ such that for any $n\ge N,$ we have $0 < a^n < \frac{\varepsilon}{2M}$. Hence for $n\ge N,$ we have \begin{align*} |u_n|\le M \frac{\varepsilon}{2M}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} Thus $u_n\to 0$ as $n\to +\infty$.

4) In this question we work with a general continuous function $f:[0,1]\to \mathbb{R}$. We set $g(t)=f(t)-f(1)$. The function $g$ is continuous on $[0,1]$ and satisfies $g(1)=0$. According to the first question we have \begin{align*} \lim_{n\to+\infty} n\int^1_0 t^n g(t)dt=0. \end{align*} In addition, observe that \begin{align*} u_n&=n\int^1_0 t^n g(t)dt+n f(1) \int^1_0 t^n dt \cr &= n\int^1_0 t^n g(t)dt+\frac{n}{n+1}f(1) \end{align*} Hence $u_n\to f(1)$ as $n\to +\infty$.

Exercise: Let $f:\mathbb{R}\to :\mathbb{R}$ be a continuous function and let $\omega\in\mathbb{R}^\ast$. Define the following function \begin{align*} \varphi(x)=\frac{1}{\omega}\int^x_0 \sin\left(\omega (x-t)\right)f(t)\,dt,\qquad x\in\mathbb{R}. \end{align*}

  • Prove that there exist two functions $\Phi,\Psi:\mathbb{R}\to :\mathbb{R}$ such that for any $x\in\mathbb{R},$ \begin{align*} \varphi(x)=\frac{\sin(\omega x)}{\omega} \Phi(x)- \frac{\cos(\omega x)}{\omega}\Psi(x). \end{align*}
  • Show that $\varphi$ is twice differentiable on $\mathbb{R}$ and that \begin{align*} \varphi''+\omega^2 \varphi=f. \end{align*}

Solution: 1) Let $x\in\mathbb{R}$ and $t\in\mathbb{R}$. We know from trigonometric formulas that \begin{align*} \sin(\omega x-\omega t)=\sin(\omega x)\cos(\omega t)-\cos(\omega x)\sin(\omega t). \end{align*} Then \begin{align*} \varphi(x)&= \frac{\sin(\omega x)}{\omega} \int^x_0 \cos(\omega t)f(t)dt-\frac{\cos(\omega x)}{\omega} \int^x_0 \sin(\omega t)f(t)dt. \end{align*} Thus it suffices to select \begin{align*} \Phi(x)= \int^x_0 \cos(\omega t)f(t)dt\quad\text{and} \quad \Psi(x)=\int^x_0 \sin(\omega t)f(t)dt. \end{align*}

2) The functions $\Phi$ and $\Psi$ are primitives of continuous functions. Thus $\Phi$ and $\Psi$ are $C^1$ functions on $\mathbb{R}$ and that \begin{align*} \Phi'(x)=\cos(\omega x)f(x),\qquad \Psi(x)=\sin(\omega x)f(x) \end{align*} for any $x\in \mathbb{R}$. This proves that the function $\varphi$ is differential on $\mathbb{R}$ as product and sum of differentiable functions. Moreover, \begin{align*} \omega\varphi'(x)&= \omega \cos(\omega x) \Phi(x)+\sin(\omega x) \cos(\omega x)f(x)\cr & \hspace{2cm}+\omega\sin(\omega x)\Psi(x)-\cos(\omega x)\sin(\omega x)f(x)\cr &= \omega \cos(\omega x) \Phi(x)+\omega\sin(\omega x)\Psi(x). \end{align*} As $\omega\neq 0,$ then \begin{align*} \varphi'(x)=\cos(\omega x) \Phi(x)+\sin(\omega x)\Psi(x) \end{align*} This last formula show that $\varphi'$ is differentiable as product and sum of differentiable functions. Hence $\varphi$ is twice differentiable and \begin{align*} \varphi''(x)&=-\omega\sin(\omega x) \Phi(x)+\cos^2(\omega x)f(x)\cr & \hspace{2.5cm}+\omega\cos(\omega x)\Psi(x)+\sin^2(\omega x)f(x)\cr &= -\omega (\sin(\omega x) \Phi(x)-\cos(\omega x)\Psi(x))+f(x)\cr &= -\omega^2 \varphi(x)+f(x) \end{align*} for any $x\in\mathbb{R}$. This proves that \begin{align*} \varphi''+\omega^2 \varphi=f. \end{align*}

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