# You need to prove that question displaystyle{int_{{0}}^{{1}}} sin{{left(pi{m}{x}right)}} sin{{left(pi{n}{x}right)}}{left.{d}{x}right.}={leftlbrace{left(begin{matrix}{0}{m}ne{n}{1}text{/}{2}{m}={n}end{matrix}right)}right.}

Question
Integrals
You need to prove that question
$$\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\left\lbrace{\left(\begin{matrix}{0}{m}\ne{n}\\{1}\text{/}{2}{m}={n}\end{matrix}\right)}\right.}$$

2021-01-20
We have to show
$$\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\left\lbrace{\left(\begin{matrix}{0}{m}\ne{n}\\{1}\text{/}{2}{m}={n}\end{matrix}\right)}\right.}$$
We will use the following trigonometric identities:
$$\displaystyle \sin{{A}} \sin{{B}}=\frac{1}{{2}}{\left[ \cos{{\left({A}-{B}\right)}}- \cos{{\left({A}+{B}\right)}}\right]}$$
Thus we have
$$\displaystyle \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}=\frac{1}{{2}}{\left[ \cos{{\left({m}-{n}\right)}}\pi{x}- \cos{{\left({m}+{n}\right)}}\pi{x}\right]}$$
First consider the case $$\displaystyle{m}\ne{n}$$. Then we have
$$\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}\frac{1}{{2}}{\left[ \cos{{\left({m}-{n}\right)}}\pi{x}- \cos{{\left({m}+{n}\right)}}\pi{x}\right]}{\left.{d}{x}\right.}$$
$$\displaystyle={\left(\frac{{ \sin{{\left[{\left({m}-{n}\right)}\pi{x}\right]}}}}{{{2}{\left({m}-{n}\right)}\pi}}-\frac{{ \sin{{\left[{\left({m}+{n}\right)}\pi{x}\right]}}}}{{{2}{\left({m}+{n}\right)}\pi}}\right)}{{|}_{{0}}^{{1}}}$$
$$\displaystyle={0},$$
because $$\displaystyle \sin{{\left({k}\pi\right)}}={0}$$, for all k in ZZ. Considering $$\displaystyle{m}={n}$$ than we have
$$\displaystyle{\int_{{0}}^{{1}}}{{\sin}^{2}{\left(\pi{m}{x}\right)}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}\frac{1}{{2}}{\left[{1}- \cos{{2}}{m}\pi{x}\right]}{\left.{d}{x}\right.}$$
$$\displaystyle={\left(\frac{x}{{2}}-\frac{{ \sin{{\left[{2}{m}\pi{x}\right]}}}}{{{4}{m}\pi}}\right)}{{|}_{{0}}^{{1}}}$$
$$\displaystyle=\frac{1}{{2}},$$
And we get the finally answer is
$$\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\left\lbrace{\left(\begin{matrix}{0}{m}\ne{n}\\{1}\text{/}{2}{m}={n}\end{matrix}\right)}\right.}$$

### Relevant Questions

a) If $$\displaystyle f{{\left({t}\right)}}={t}^{m}{\quad\text{and}\quad} g{{\left({t}\right)}}={t}^{n}$$, where m and n are positive integers. show that $$\displaystyle{f}\ast{g}={t}^{{{m}+{n}+{1}}}{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}$$
b) Use the convolution theorem to show that
$$\displaystyle{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}=\frac{{{m}!{n}!}}{{{\left({m}+{n}+{1}\right)}!}}$$
c) Extend the result of part b to the case where m and n are positive numbers but not necessarily integers.
Give the correct answer and solve the given equation
Let $$\displaystyle{p}{\left({x}\right)}={2}+{x}{\quad\text{and}\quad}{q}{\left({x}\right)}={x}$$. Using the inner product $$\langle\ p,\ q\rangle=\int_{-1}^{1}pqdx$$ find all polynomials $$\displaystyle{r}{\left({x}\right)}={a}+{b}{x}\in{P}{1}{\left({R}\right)}{P}$$
(R) such that {p(x), q(x), r(x)} is an orthogonal set.
Give the correct answer and solve the given equation $$\displaystyle{\left({x}-{y}\right)}{\left.{d}{x}\right.}+{\left({3}{x}+{y}\right)}{\left.{d}{y}\right.}={0},\text{when}\ {x}={3},{y}=-{2}$$
Let $$\displaystyle\le{f}{t}{\left\lbrace{v}_{{{1}}},\ {v}_{{{2}}},\dot{{s}},\ {v}_{{{n}}}{r}{i}{g}{h}{t}\right\rbrace}$$ be a basis for a vector space V. Prove that if a linear transformation $$\displaystyle{T}\ :\ {V}\rightarrow\ {V}$$ satisfies $$\displaystyle{T}{\left({v}_{{{1}}}\right)}={0}\ \text{for}\ {i}={1},\ {2},\dot{{s}},\ {n},$$ then T is the zero transformation.
Getting Started: To prove that T is the zero transformation, you need to show that $$\displaystyle{T}{\left({v}\right)}={0}$$ for every vector v in V.
(i) Let v be an arbitrary vector in V such that $$\displaystyle{v}={c}_{{{1}}}\ {v}_{{{1}}}\ +\ {c}_{{{2}}}\ {v}_{{{2}}}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}\ {v}_{{{n}}}.$$
(ii) Use the definition and properties of linear transformations to rewrite $$\displaystyle{T}\ {\left({v}\right)}$$ as a linear combination of $$\displaystyle{T}\ {\left({v}_{{{1}}}\right)}$$.
(iii) Use the fact that $$\displaystyle{T}\ {\left({v}_{{i}}\right)}={0}$$ to conclude that $$\displaystyle{T}\ {\left({v}\right)}={0}$$, making T the zero tranformation.
Give the correct answer and solve the given equation
$$\displaystyle{\left({x}+{y}\right)}{\left.{d}{x}\right.}+{\left({x}-{y}\right)}{\left.{d}{y}\right.}={0}$$
Let f(t) be a function on $$\displaystyle{\left[{0},\infty\right)}$$. The Laplace transform of fis the function F defined by the integral $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$ . Use this definition to determine the Laplace transform of the following function.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$
A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
Give the correct answer and solve the given equation Evaluate $$\displaystyle\int{x}^{3}{\left({\sqrt[{3}]{{{1}-{x}^{2}}}}\right)}{\left.{d}{x}\right.}$$
Give the correct answer and solve the given equation $$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{y}^{2}-{1}}}{{{x}^{2}-{1}}},{y}{\left({2}\right)}={2}$$
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$