Math 8 Practice Exam Problems - solutions

Surely, the only errors are typographical. Well, probably...

1.

Evaluate the integral $$\int (\sin x + x)^2\, dx$$.

Solution: $$\int (\sin x + x)^2\, dx = \int (\sin^2 x + 2x\sin x + x^2)\,dx$$.

$$\int \sin^2x\,dx = \int\frac{1-\cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$$.

$$\int 2x\sin x\,dx = -2x\cos x + 2\sin x + C$$ (by parts u = 2x).

Putting things togther we get
$$\int (\sin x + x)^2\, dx = \frac{x}{2} - \frac{\sin 2x}{4} -2x\cos x + 2\sin x + \frac{x^3}{3} + C$$.

2.

Find the general solution to the differential equation $$(D-3)^4(D^2 + 4D + 5)^2 y = 0$$.

Solution: The auxilliary equation is (r-3)⁴(r² + 4r + 5)² which has real root 3 (with multiplicity 4) and complex conjugate roots $-2\pm i$ (with multiplicity 2).

Thus the general solution will be of the form:
$C_1e^{3x} + C_2xe^{3x} + C_3x^2e^{3x} + C_4x^3e^{3x} + C_5e^{-2x}\cos x + C_6e^{-2x}\sin x + C_7xe^{-2x}\cos x + C_8xe^{-2x}\sin x$.

3.

Consider the homogeneous differential equation $$\frac{d^2y}{dt^2} + 9 y = 0$$. Find the general solution as well as the particular solution which satisfies the initial conditions y(0) = 3 and y'(0)= 6.

Solution: The auxilliary equation is r² + 9 = 0 which as roots $\pm3i$. The general solution has the form $y = A\cos 3t + B\sin 3t$. y(0) = 3 implies A = 3. $y'(t) = -3A\sin 3t + 3B\cos 3t$, so y'(0) = 6 implies B=2, so the particular solution is $y = 3\cos 3t + 2\sin 3t$.

4.

Find the general solution to the differential equation $$\frac{dy}{dx} = \frac{\ln x + 1}{y}$$, which passes through the point $(1,2\sqrt2)$.

Solution: This is separable and yields:

$$\int y\,dy = \int (\ln x + 1)\,dx$$. Integrating the log by parts ($u = \ln x$) yields $y^2/2 = x\ln x + C$. The initial condition yields C = 4, so the particular solution is $y = \sqrt{2x\ln x + 8}$. We note that the solution $y = -\sqrt{2x\ln x + 8}$ passes through the point $(1,-2\sqrt2)$.

5.

Consider the differential equation $$\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0$$, where b and c are constants and $b = 2\sqrt c$. Determine the general solution of this differential equation.

Solution: The auxilliary equation is r² + br + c = 0 which has the double real root $$r = \frac{-b \pm \sqrt{b^2 - 4c}}{2} = \frac{-b}{2}$$.

Thus the general solution is y = C₁e^-bx/2 + C₂xe^-bx/2.

6.

Find the volume of the solid of revolution obtained by revolving the region bounded by the curves y = e^x, y = 0, x=0, and x=1 about the line x = 3.

Solution: The volume (shells) is given by

$$\int_0^1 2\pi (3 - x) e^x \, dx = 2\pi(3e - 4)$$ via integration by parts.

7.

The rate at which students forget formulas is proportional to the number of formulas ( $$\frac{dF}{dt} = kF$$) If 10 minutes after walking into an exam, the student remembers only a third of the formulas initially stored in short-term memory, what fraction of the formulas are remembered 30 minutes into the exam.

Solution: $$\frac{dF}{dt} = kF$$ has the general solution $$F(t) = F(0)e^{kt}$$ which you can find by separating variables. F(10) = F(0)/3 so $$F(0)e^{10k} = F(0)/3$$. We are assuming F(0) > 0 (hopefully a valid assumption). Then $k = -\ln3 / 10$, and $$F(t) = F(0)e^{-tln 3/10}$$, so $$F(30) = F(0)e^{-3ln 3} = F(0)/27$$, thus after 30 minutes 1/27th of the formulas are remembered.

8.

Derive a reduction formula for $\int\sin^n x\,dx$ for $n \ge 2$. That is derive an expression for $\int\sin^n x\,dx$ in terms of $\int\sin^m x\,dx$ for m < n.

Solution: $$\int\sin^n x\,dx = \int \sin^{n-1}x \sin x\,dx$$ Using $u = \sin^{n-1}x$ and $dv = \sin x\,dx$ yields $du = (n-1)\sin^{n-2}x\cos x$ and $v = -\cos x$. Integration by parts yields

$$\int\sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\cos^2 x\, dx$$.

Since $\cos^2 x = 1 - \sin^2 x$, we substitute to get:

$$\int\sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int (\sin^{n-2}x - \sin^n x)\,dx$$.

Shifting the $(n-1)\int \sin^n x\,dx$ to the left-hand side, we get:

$$n\int\sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\,dx$$, or

$$\int\sin^n x\,dx = -\frac{1}{n}\sin^{n-1}x\cos x + \frac{(n-1)}{n}\int \sin^{n-2}x\,dx$$