int_{8}^{21}f(x)dx-int_{8}^{11}f(x)dx=int_{a}^{b}f(x)dx

Question
Integrals
$$\int_{8}^{21}f(x)dx-\int_{8}^{11}f(x)dx=\int_{a}^{b}f(x)dx$$

2021-02-09
$$\int_{8}^{21}f(x)dx-\int_{8}^{11}f(x)dx=\int_{a}^{b}f(x)dx=\int_{8}^{11}f(x)dx+\int_{8}^{21}f(x)dx=\int_{11}^{21}f(x)dx$$
So a=11, b=21

Relevant Questions

$$\int_{b}^{a}x^{7}dx$$
$$\displaystyle{\int_{{{b}}}^{{{a}}}}{x}^{{{7}}}{\left.{d}{x}\right.}$$
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.
$$\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}$$
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.
$$\int_{1/8}^1\frac{dx}{x\sqrt{1+x^{2/3}}}$$
$$\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\left\lbrace{\left(\begin{matrix}{0}{m}\ne{n}\\{1}\text{/}{2}{m}={n}\end{matrix}\right)}\right.}$$
Evaluate the integral $$\int \frac{1}{1+\frac{x}{2}^2}dx$$
Evaluate the integral $$\displaystyle\int{\frac{{{1}}}{{{1}+{\frac{{{x}}}{{{2}}}}^{{2}}}}}{\left.{d}{x}\right.}$$
Evaluate the integral. $$\displaystyle\int{x}\sqrt{{{5}{x}-{1}}}{\left.{d}{x}\right.}$$