Search Results

It sounds like what you did was to write $$ \int_0^\infty \left( \sin(1/x) - \frac{\sin(\pi/x)}{\pi} \right) \,dx = \int_0^\infty \sin(1/x)\, dx - \int_0^\infty \frac{\sin(\pi/x)}{\pi} \,dx $$ You the …

Just as another way to see it, you could rewrite the relationship between $u$ and $r$ as $r = \frac{4}{3} u$ and $dr = \frac{4}{3} du$. Then you would have $$ \frac7{16}\int\frac1{1+\frac{9}{16}r^2}d …

Begin by considering an arbitrary (non-zero) constant vector $\vec{c}$ alongside our vector field $\vec{v}$. We have $$ \vec{\nabla} \cdot (\vec{v} \times \vec{c}) = \vec{c} \cdot (\vec{\nabla} \tim …

This is the sort of problem that the calculus of variations was invented to solve. …

To elaborate on my comment: Let's look for a solution of the form $f(x) = C e^{\alpha x}$ for some (possibly complex) constants $\alpha$ and $C$. Then we have \begin{align*} \frac{d}{dx} \left[ \lef …

For part (a), you're on the right track; note, however, that the equation is just as well-defined for $y < -1$ as it is for $y > -1$. @crankk's suggestion to apply the Picard-Lindelöf Theorem would …

Promoting from and elaborating on a comment, because it seems to have resolved the OP's confusion: It would make sense if you wrote $$ f(x_0+g(x))=f(x_0)+f^{\prime}(x_0)g(x)+\frac{1}{2}f^{\prime\prim …

If all you know is that the line $r$ lies in the desired plane, then the plane is not unique. If we find one such plane, we can rotate it about the line $r$ to obtain another plane in which $r$ lies. …

I think that $$ a_n = a_1 - 7 b_1 + 4(n-1) + 7 \left\lfloor \frac{6(n+b_1)}{7} \right\rfloor $$ will do the trick. Here, $a_1$ is the first number in the sequence, $b_1 = a_1 \bmod 10$ is the last di …

The area between the parabola and this diagonal can be found by standard Calculus 101 techniques. …

The first term of the integral is odd in $x$, and the annulus is symmetric under the reflection $x \to -x$. This implies that it integrates to zero. A similar argument using the $y$-direction shows …

Imagine slicing the cone into circular cross-sections. These cross-sections get bigger as you go up the cone. Let's think about two cross-sections in particular: one slice spanning between 1 cm and …

While this is probably not the intended method to solve this problem, you can actually sum this series up explicitly via a quick detour into the complex plane and examine the properties of the resulti …

Stirling's approximation says that $$ \ln \left[ x! e^{-x^2} \right] = \ln x! - x^2 \approx (x \ln x - x) - x^2 $$ and since this quantity goes to $-\infty$ as $x \to \infty$, we conclude that $$ \lim …

A typical function that models this sort of behavior is a logistic function: $$ f(x) = \frac{1}{1 + e^{-(x - a)/d}} $$ where $a$ is the "center" of the function (such that $f(a) = \frac12$), and $d$ i …