# Tag Info

## New answers tagged definite-integrals

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So, you know the two curves intersect at $p = \cos^{-1} (1/2)$, which implies $p = \pm \pi/3 + 2\pi k$ for any $k \in \mathbb{Z}$. For which $k$ do we observe an intersection within the specified integral of integration? That is, for which $k$ is $p \in [0, \pi/7]$? (Hint: there is only one such $k$) The last step is to integrate $f(x) - g(x)$ from $0$ to ...

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As was mentioned: the integrand simplifies to: $\sqrt{a^2 + b^2 - 2 a b \cos\theta \sin\phi}$ Factor out $\sqrt{a^2 +b^2}$ and use the binomial theorem to expand, treating $\cos\theta \sin \phi$ as small. This gives: $$\sqrt{a^2 +b^2} \left(1-\frac{a b \cos (\theta ) \sin (\phi )}{a^2+b^2}-\frac{a^2 b^2 \cos ^2(\theta ) \sin ^2(\phi )}{2 ... 3 We will use similar approach as sos440's answer in I&S. Using the simple algebraic identity$$ ab^2=\frac{(a+b)^3+(a-b)^3-2a^3}{6}, it follows that \begin{align} \int_0^1 \frac{\ln x\ln(1+x)\ln^2(1-x)}{x}\ dx &=\frac16I_1+\frac16I_2-\frac13I_3\ ,\tag1 \end{align} where \begin{align} I_1&=\int_0^1\frac{\ln x\ln^3(1-x^2)}{x}\ dx\\[12pt] ... 2I=\int_{-2\pi}^{2\pi} xe^{-\mid x\mid}dx.$$Loosely speaking, you can think about the definite integral as the area bounded by the function xe^{-\mid x\mid} and the x-axis, as the variable x moves from x=-2\pi to x=0 then from x=0 through to x=2\pi. So, intuitively it's not too much of a step to see that$$I=\int_{-2\pi}^{0} xe^{-\mid ...

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It depends on your definition/understanding of "definite integral". If you think of the Lebesgue definite integral, the answer is yes, because each function get mapped to a scalar value and this action is linear. However, this action may not be bounded. There are, however, more general ideas of definite integrals, e.g., the Bochner integral. Think of a ...

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