Enable contrast version

# Tutor profile: Rachael V.

Inactive
Rachael V.
Math student
Tutor Satisfaction Guarantee

## Questions

### Subject:Study Skills

TutorMe
Question:

I need to cram for a math exam, where do I start?

Inactive
Rachael V.

The most efficient way to cram for a math exam is to first understand any example problems or practice exam questions provided by the instructor. If none are provided or you have already gone through them, next go through past homework, prioritize medium difficulty questions, although make sure basic question are understood since those are usually building blocks needed for more difficult questions. Finally if you have extra time in studying, go over the difficult homework questions and review class notes and/or textbook.

### Subject:Calculus

TutorMe
Question:

(multivariable calculus question) Solve the following integral: $$\int_{\mathbb{R}}e^{-x^2}dx$$

Inactive
Rachael V.

This integral is complicated to solve outright, even though it looks simple, so we will use a trick. Trick: Let $$L=\int_{\mathbb{R}}e^{-x^2}dx$$ and notice that $$\int\int_{\mathbb{R}}e^{-x^2-y^2}dxdy=\int_{\mathbb{R}}e^{-x^2}dx\int_{\mathbb{R}}e^{-y^2}dy=L^2$$ Step 1: Solve $$\int\int_{\mathbb{R}}e^{-x^2-y^2}dxdy$$ Step 2: Apply polar coordinates, $$x=r\cos(\theta), y=r\sin(\theta)$$ $$-x^2-y^2=-r^2\cos^2(\theta)-r^2\sin^2(\theta)=-r^2(\cos^2(\theta)+\sin^2(\theta))$$ Apply trig identity, $$\cos^2(\theta)+\sin^2(\theta)=1$$ $$-x^2-y^2=-r^2$$ $$dxdy=rd\theta dr$$ Step 3: Plug in polar coordinates into integral $$\int\int_{\mathbb{R}}e^{-x^2-y^2}dxdy=\int_{0}^{\infty}\int_{0}^{2\pi}e^{-r^2}rd\theta dr=2\pi\int_{0}^{\infty}e^{-r^2}rdr$$ Step 4: Apply u-substitution $$u=r^2, du=2rdr$$ $$2\pi\int_{0}^{\infty}e^{-r^2}rdr=2\pi\int_{0}^{\infty}\frac{1}{2}e^{-u}du=\pi\int_{0}^{\infty}e^{-u}du$$ Note that $$e^0=1$$ and $$e^{-\infty}=\frac{1}{e^{\infty}}=0$$ $$\pi\int_{0}^{\infty}e^{-u}du=\pi$$ Step 5: Apply Trick $$L^2=\pi, L=\sqrt{\pi}$$ Solution: $$\int_{\mathbb{R}}e^{-x^2}dx=\sqrt{\pi}$$

### Subject:Algebra

TutorMe
Question:

Find all solutions: $$x^3+7+2x^2+4x=6x^2+7$$

Inactive
Rachael V.

Step 1: Move all terms to the left side of equation $$x^3+7+2x^2+4x-6x^2-7=0$$ Step 2: Group like terms $$x^3-4x^2+4x=0$$ Step 3: Factor $$x(x^2-4x+4)=0$$ $$x(x-2)(x-2)=0$$ Step 4: Solution $$x=0,2$$ Alternate method: In step 4 instead of factoring fully or if unable to factor fully can also use the quadratic formula after factoring out the $$x$$, so we still have $$x=0$$ for our first solution and then using the quadratic formula to find the other solution(s). Take $$a=1, b=-4, c=4$$ $$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}=\frac{+4+/-\sqrt{(-4)^2-4(1)(4)}}{2(1)}=\frac{4+\sqrt{16-16}}{2},\frac{4-\sqrt{16-16}}{2}=2,2$$ So using our alternate method we obtain the same solutions for $$x$$, $$x=0,2$$.

## Contact tutor

Send a message explaining your
needs and Rachael will reply soon.
ContactÂ Rachael