Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [integration]

For questions about the properties of integrals. Use in conjunction with (indefinite-integral), (definite-integral), (improper-integrals) or another tag(s) that describe the type of integral being considered. This tag often goes along with the (calculus) tag.

Integration is a major part of .

There are two main kinds of integrals:

  • definite integrals (e.g. proper and improper integrals), which often have numerical values
  • indefinite integrals, which group families of functions with the same derivative.

Several techniques to solve integrals have been developed, including integration by parts, substitution, trigonometric substitution, and partial fractions.

Integration can be used to find the area under a graph and find the average of the function. Also, it can be used to compute the volume of certain solids and to find the displacement of a particle.

73636 questions
3
votes
1 answer

Trouble with an integral

I've difficulties calculating the following sum of integrals: $$\int_{0}^{\sqrt2}{{\int_{0}^{x}xe^{{(x^2+y^2)}^{3/2}}dy} dx}+\int_{\sqrt2}^{2}{{\int_{0}^{\sqrt{4-x^2}}xe^{{(x^2+y^2)}^{3/2}}dy} dx}$$ My suggestion would be to rewrite this…
Bart Rutten
  • 78
  • 6
3
votes
3 answers

integral of derivated function mix

$$\int_0^{2\pi}(t-\sin t){\sqrt{1-\cos t}} dt$$ I can notice that I have something of the form $$\int{f(x){\sqrt {f'(x)}}dx}$$ but I don't know anything that could simplify it
Jonny M
  • 95
  • 6
3
votes
2 answers

Evaluate integral with exponential and polynomial

How can I show that \begin{align} \int_{-\infty}^\infty e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx=\infty \end{align} for $t\neq0$. I started as follows: \begin{align} \int_{-\infty}^\infty e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx=\int_{-\infty}^0…
Sha Vuklia
  • 3,960
  • 4
  • 19
  • 37
3
votes
2 answers

Can a function be an antiderivative? $u=\int\frac{\mathrm du}{\mathrm dx}\mathrm dx$

Why can we say the following: $u=\int\frac{\mathrm du}{\mathrm dx}\mathrm dx$, where we treat $u$ as a function (e.g., see this proof). Because as far as I know, $\int\frac{\mathrm du}{\mathrm dx}\mathrm dx$, as an indefinite integral, stands for a…
Sha Vuklia
  • 3,960
  • 4
  • 19
  • 37
3
votes
2 answers

Solving $\int \frac{1}{\sqrt{1-a+ax^2}}dx $

This is part of a large engineering report and through plugging this into wolfram alpha I have an answer that, once computed, works perfectly for what I need. However, I am struggling to see how to go from the input to the output, if anyone could…
M123
  • 31
  • 2
3
votes
0 answers

Evaluate $\int \frac{x+1}{5x+4}dx$

$$\int \frac{x+1}{5x+4}dx$$ $$\frac{1}{5}\int \frac{5(x+1)}{5x+4}dx=\frac{1}{5}\int \frac{5x+5}{5x+4}dx=\frac{1}{5}\int (\frac{5x+4}{5x+4}+\frac{1}{5x+4})dx=\frac{1}{5}\int (1+\frac{1}{5x+4})dx=\frac{1}{5}…
gbox
  • 12,867
3
votes
2 answers

$\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x+1}dx$

What results? $$\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x+1}dx$$ my try : $u= \tan \frac{x}{2 } $ but : What is the short way?
Almot1960
  • 4,782
  • 16
  • 38
3
votes
0 answers

Finding irrational function including $\sin$ and $\cos$

Finding $\displaystyle \int \frac{\cos^4 x-\sin^4 x}{\sqrt{1+\cos^4 x}}dx$ Attempt let $\displaystyle I = \int\frac{(\cos^2 x-\sin^2 x)(\cos^2 x+\sin^2 x)}{\sqrt{1+\cos^4 x}}dx = \int\frac{\cos 2x}{\sqrt{1+\cos^4 x}}dx$ wan,t be able to go further,…
DXT
  • 11,241
3
votes
2 answers

$f(x)$ is continuous on $[0,\pi]$.Prove that the following limit is $\frac{2}{\pi}\int_{0}^{\pi}f(x)$

$f(x)$ is continuous on $[0,\pi]$.Prove that $$\lim_{n\to\infty}\int_0^\pi |\sin(nx)|f(x)\ dx = \frac 2\pi \int_0^\pi f(x) dx$$ I made the substitution $x=\frac{u}{n} \implies \text{dx}=\frac{1}{n} \text{du}$. So, we have $\lim_{n \rightarrow…
user321656
3
votes
4 answers

Integrate $\int r(\ln r)^2 \, dr$

I'm trying to integrate $\int r(\ln r)^2\, dr$. I let $u=\ln r$ and $du=\dfrac 1 r \, dr$. So then I get, $\displaystyle\int \dfrac{ru^2} r \, du = \int u^2 \, du = \dfrac 1 3 (u)^3+C$$$=\frac 1 3 (\ln r)^3 +C$$ Now wolfram alpha does the problem…
Ryan Tandy
  • 403
3
votes
1 answer

Definite integral of a combination of exponents

Is there a closed-form formula (e.g. in terms of some known special functions) of $$\int_{-\infty}^{\infty}\frac{\exp(-x^2)}{1+b\exp(-\alpha x)}d x$$
Anton
  • 33
  • 3
3
votes
1 answer

Evaluate $\int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x$

How to evaluate $$\displaystyle \int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x$$ I tried $$\int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x=\frac{1}{3}\int_{0}^{1}x^{-\frac{1}{3}}\left ( 1+x \right )^{\frac{1}{2}}\mathrm{d}x$$ But I can't figure out how to go on.
user220660
3
votes
2 answers

Two integrable function such that $\left( \int_1^3 f(x) g(x) dx \right)^2 = \left( \int_1^3 f^2(x) dx \right) \left( \int_1^3 g^2(x) dx \right)$

Let $f(x)$ and $g(x)$ are two integrable function, $x \in [1,3]$ satisfying $$\left( \int_1^3 f(x) g(x) dx \right)^2 = \left( \int_1^3 f^2(x) dx \right) \left( \int_1^3 g^2(x) dx \right)$$ Given that f(1)=2 and g(1)=4 My sir told me…
user135482
  • 139
3
votes
5 answers

How to evaluate integral: $\int x^2\sqrt{x^2+1}\;dx.$

I want to solve the integral:$$\int x^2\sqrt{x^2+1}\;dx.$$ I did $x = \tan t$, then it is equal to:$$\int\frac{\tan^2 t}{\cos^3 t}\;dx.$$ Or:$$\int\frac{\sin^2 t}{\cos^5 t}\;dx.$$ I stuck there. Any help will be much appreciated...
Olimjon
  • 607
3
votes
1 answer

Weierstrass Substitution

I was reading up about the Weierstrass Substitution and don't understand what 'No generality is lost' means in this context. In integral calculus, the tangent half-angle substitution is a substitution used for finding antiderivatives, and hence…
user401334
  • 115
1 2 3
…
99 100