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Questions tagged [definite-integrals]

Questions about the evaluation of specific definite integrals.

A definite integral is defined as the area under a function from $a$ to $b$. Definite integrals sometimes involve calculating the indefinite integral, which is a function giving the area from $0$ to any $x$. However, definite integrals are most often separate from indefinite integrals in that the indefinite integral may not exist on its own. This is usually in the case of piece wise functions that are split along certain key points or integrals involving asymptotes.

20559 questions
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Gauss' divergence theorem with a bounded parabaloid

I've been given a region bounded by the values $z = x^2 + y^2$ and $z = 1$, with a force of $F = 2i + xj + z^3k$. So with Gauss' theorem I have been able to deduce: $$\int\int_S -F1f_x - F2f_y + F3dS = \int\int\int_V \frac{\delta}{\delta{x}}F1 +…
Laefica
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Closed-form expression for integral involving exponential of cosines?

I'm looking for a closed-form expression for the following integral: $$ \int_0^{2\,\pi} \exp\{a_1\,\cos(\theta_1-\mu) + a_2\,\cos(\theta_2 - \mu)\}\,d\mu $$ where $a_1,a_2 > 0$ and $0 \le \theta_1, \theta_2 < 2\,\pi$.
mef
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$g(2-x)=g(2+x)$ and $g(4-x) = g(4+x)$ and $ \int^{2}_{0}g(x)dx = 2.$ then $\int^{50}_{0}g(x)dx$ is

function $g(x)$satisfy following condition $g(2-x)=g(2+x)$ and $g(4-x) = g(4+x)$ and $\displaystyle \int^{2}_{0}g(x)dx = 2.$ then $\displaystyle \int^{50}_{0}g(x)dx$ is from $g(2-x) = g(2+x)$ $g(2-(2-x)) = g(2+2-x)$ $g(x) =…
DXT
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How to solve $\int_0^{2\pi} $ $(\int_1^\infty$ $\sin(s-t) \over t^3$ $dt$ $)ds$?

I already tried to solve it myself and to type it into the calculator, but it gives out some weird stuff with an endless long solution.
Julian
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Integral that is not defined at upper bound

I want to evaluate the following integral: $$ \int_{1/2}^1 \frac{1}{x \sqrt{(1-x)x}} dx$$ I would compute this as $$ \frac{2(x-1)}{\sqrt{(1-x)x}}\biggr\rvert_{1/2}^1 $$ which is not defined at $1$. However, Mathematica evaluates the definite…
bonifaz
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Integral involving exponential of inverse fourth power of sine

I need to solve the following integral: \begin{align*} \int_{0}^{\pi} \sin^2(\theta) e^{-\frac{c}{\sin^4 \theta}} d\theta \end{align*} I tried a few changes of variables to make it look like a Gaussian integral, as well as integration by parts, but…
karakusc
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Definite integral of sinc function to power two and three

Consider definite integral of sinc function to power $p$ $I(p) = \int_0^{\infty} = \left(\frac{\sin(x)}{x}\right)^p dx$ For simplest case $p=1$, well known result is $I(1) = \pi/2$. How to derive result for $p=2$ and $p=3$?
Nigel1
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Evaluation of $\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx$

Evaluation of $$\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx$$ Let $$I = \int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx = \int^{1}_{0}\frac{(1-x)+\ln x}{(1-x)\ln x}dx$$ Now How can i solve after that , Help required, Thanks
juantheron
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How do I calculate value of integral?

How do I calculate value of integral? It's given interval $I=[0,5] \times [0,6]$ how do I calculate $\int_{I} −5xy^3 \; d(x,y)$ ? What is confusing me is that I don't know should I calculate $\int_{0}^{5}\int_{0}^{6} −5xy^3 \; dx \; dy$? or ?
Alen
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Relation of two definite integrals

For a differentiable continuous function $g(u)$ over a finite region $R$, I know that $$\int_R e^u g\; u\; du = 1$$ Is there any way to determine from this the value of $$\int_R 10^u g\; u\; du$$
Oliver
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Evaluation of $\int^{1}_{0}\frac{\ln(13-6x)}{\sqrt{1-x^2}}dx$

Evaluation of $$\int^{1}_{0}\frac{\ln(13-6x)}{\sqrt{1-x^2}}dx$$ $\bf{My\; Try::}$ Put $x=\cos \theta\;,$ Then $dx = -\sin \theta d\theta$ and changing limits, We get $$I = -\int^{0}_{\frac{\pi}{2}}\frac{\ln(13-6\cos \theta)}{\sin \theta}\cdot \sin…
juantheron
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$\int _2^3\:\frac{x^2-2x-3}{x+1}dx$

After I get form of $\int \frac{x^2}{x+1}dx-\int \frac{2x}{x+1}dx-\int \frac{3}{x+1}dx$ then get $\left(x+1\right)^2-2\left(x+1\right)+ln\left|x+1\right|-2\left(x+1-ln\left|x+1\right|\right)-3ln\left|x+1\right|$ (problem here i think ^) using…
user372522
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$\int _0^1\:2xe^{x^2-1}dx$

for the answer I'm getting: $1+\frac{1}{e}$ but according to symbolab, I get: $1-\frac{1}{e}$ steps I'm currently taking with u substitution; I get: $\int \:e^u$ then plugging back u; $e^{x^2-1}$ (from 0 to 1) then I compute, $e^{-1}+e^0$ =…
user372522
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$\int _0^2\:\frac{2x-1}{\:x^2-3x+2}dx$

how am I able to solve this definite integral when it goes from 0 to 2; I know how to solve it for example $\int _3^5\frac{2x-1}{x^2-3x+2}dx$ but not when limit is approaching from a negative number or a zero... $\int…
user372522
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Evaluation of $\int^{2}_{0}\frac{1}{\sqrt{1+x^4}}dx$

Let $$I = \int\frac{1}{\sqrt{1+x^4}}dx\;,$$ Now put $x^2=\tan t\;,$ Then $2xdx = \sec^2 tdt$ So $$I = \frac{1}{2}\int\frac{\sec^2 t}{\sec t \sqrt{\tan t}}dt = \frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{\sin 2 t}}dt$$ Now put $\displaystyle…
juantheron
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