# Tag Info

0

Leave out $1/x^2$, to be reinserted later: \begin{align} \int xe^x\,dx &=\sum_{k=0}^{\infty}\frac{x^{k+2}}{(k+2) k!}\\ &=\sum_{k=0}^{\infty}\frac{(k+1)x^{k+2}}{(k+2)!}\\ &=\sum_{k=2}^{\infty}\frac{(k-1)x^k}{k!}\\ &=\sum_{k=2}^{\infty}\frac{kx^k}{k!}-\sum_{k=2}^{\infty}\frac{x^k}{k!}\\ ...

6

We have $$\int_0^t e^{\lambda x}dx = \dfrac{e^{\lambda t}-1}{\lambda}$$ Differentiating with respect to $\lambda$, we obtain $$\int_0^t xe^{\lambda x}dx = \dfrac{\lambda t e^{\lambda t} - e^{\lambda t }+1}{\lambda^2}$$ Set $\lambda = 1$, to obtain $$\int_0^t xe^{x}dx = t e^{t} - e^{t }+1$$ EDIT To complete your approach, note that \sum_{k=0}^{\infty} ... 0 This is an alternative inspired by the technique used in this answer. I may or may not delete this answer depending on your answer to my comment in the question. Take the ansatz \displaystyle \int \cos(x)e^{-x}\mathrm dx=(A\cos(x)+B\sin(x))e^{-x} with A,B yet to be determined. Differentiating both sides yields, for all x\in \mathbb R, ... 0 Letting u = e^{-x} and dv = \cos x \, dx so that du = -e^{-x} \, dx and v = \sin x, we obtain: \begin{align*} \int e^{-x}\cos x \, dx &= (e^{-x})(\sin x) - \int (\sin x)(-e^{-x} \, dx) \\ &= e^{-x}\sin x + \int e^{-x}\sin x \, dx \end{align*} Letting u = e^{-x} and dv = \sin x \, dx so that du = -e^{-x} \, dx and v = -\cos x, we ... 0 Using e^{-x}=dv and cos(x)=u:-e^{-x}cos(x)-\int e^{-x}sin(x)\int e^{-x}sin(x)=-e^{-x}sin(x)-\int -e^{-x}cos(x)$$Be careful with the signs, and you will get an expression like this:$$2\int e^{-x}cos(x)=-e^{-x}cos(x)+e^{-x}sin(x)$$Divide by two, and that is your solution. 2 First, set u = \cos x and dv = e^{-x} dx, so du = - \sin x \,dx and v = - e^{-x}. We get$$\int \cos (x) e^{-x} \,dx = (\cos (x))(-e^{-x}) - \int (-\sin (x))(-e^{-x})\, dx$$. Now, set u = - \sin x and dv = -e^{-x}\,dx to get du = - \cos x\, dx  and v = e^{-x}. This gives us$$\int \cos (x) e^{-x} \,dx = (\cos(x))(-e^{-x}) - (-\sin(x))(e^{-x}) ...

1

$$\int\cos xe^{-x}dx=Re\int e^{ix}e^{-x}dx=Re\int e^{x(i-1)}dx$$

0


1

You're almost there. Let $\displaystyle h(x) = g(x) \int_a^b f(t) \ dt$. As $g$ is continuous, $h$ is also continuous. Without loss of generality, let $x_1 < x_2$. By what you've shown above, $\int_a^b f(x)g(x) \ dx$ is a number between $h(x_1)$ and $h(x_2)$. As $h$ is continuous, by the IVP there must be a value $x_0 \in (x_1, x_2)$ such that ...

3

The ML inequality is (essentially) a real inequality. It holds for all (sufficiently regular, e.g. piecewise differentiable) curves and [again, sufficiently regular so that the integral is defined] functions [or vector fields] in any $\mathbb{R}^n$, $\mathbb{C}^n$ or more generally, Riemannian manifold. Its proof uses a) the inequality for real intervals ...

1

Given only the information stated, the only reason we can assume that 800-p isn't negative is that we are taking its logarithm. This makes sense in terms of the model; 800 is functioning as a population ceiling, as the rate of increase slows down as $p$ approaches 800.

1

I think the answer is likely to be due to the fact that $800-p$ cannot be negative. I don't quite know what 800 represents but it seems likely that $800-p$ cant be a negative number otherwise it doesn't work out. Hope this makes sense. Would have added this as a comment but I cant presently.

0

We can use the following Taylor expansion of $\ln(1+x)$ to evaluate \begin{eqnarray} \ln(1+x)&=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}x^n. \end{eqnarray} Then \begin{eqnarray} \int_0^{\pi/2}\frac{1+\sin\phi}{\sin\phi}d\phi&=&\int_0^{\pi/2}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\sin^{n-1}\phi d\phi\\ ...

1

A possible contour is the semi circle in the upper half plane with a semi circle around the origin. Let $\int_{\Gamma}$ be large semi circle and $\int_{\gamma}$ be the small semi circle. Let $R$ be the radius of the $\Gamma$ and $\epsilon$ the radius of $\gamma$. Now as $R\to\infty$, $\int_{\Gamma}\to 0$ and similarly as $\epsilon\to 0$, $\int_{\gamma}\to 0$ ...

Top 50 recent answers are included