Concordia MATH 205 Midterm Practice Mock

1. A sum and a displacement (10 marks)

(a) The width is $\Delta t=2/n$ seconds; the $k$th right endpoint is $t_k=2k/n$. A velocity times this width is a displacement in metres.

$$\begin{gathered}S_n=\sum_{k=1}^{n}\left(3-\frac{4k^2}{n^2}\right)\frac2n\\=6-\frac8{n^3}\frac{n(n+1)(2n+1)}6.\end{gathered}$$

Expanding $(n+1)(2n+1)$ gives $S_n=10/3-4/n-4/(3n^2)$ metres.

(b) Thus $S_2=1\,\mathrm m$ and $\lim S_n=10/3\,\mathrm m$. Directly, $$\begin{gathered}\int_0^2(3-t^2)\,dt\\=[3t-t^3/3]_0^2=10/3\end{gathered}$$.

(c) The velocity decreases on this interval, so every right-endpoint rectangle is below the graph in the signed-sum comparison. The estimates are below the exact displacement. Since $v(t)<0$ when $\sqrt3<t\le2$, the object reverses direction. Distance is $\int_0^2|v(t)|\,dt$, which is greater than the net displacement here.

Illustrative mark guide

• (a) width and endpoints 1; correct sum 1; use of the square-sum formula and simplification 2
• (b) S₂ 1; limit 1; metres 1
• (c) decreasing-function comparison 1; reversal interval 1; distance uses absolute velocity 1

Check: The error $S_n-10/3=-4/n-4/(3n^2)$ is negative for every positive integer $n$.

D1: If you added velocities without a time width, write the units of one summand before doing algebra. If you called the signed result a distance, locate the zero of the velocity.

Guided integration practice

2. Recover a function and differentiate an accumulation (10 marks)

(a) Integrating gives $F(x)=2x^3-x^2+x+C$. The condition $F(0)=-3$ fixes $C=-3$. Thus $F(x)=2x^3-x^2+x-3$.

(b) The integrand is continuous. Write $$\begin{gathered}A(x)=\int_0^{2x}(2+t^3)\,dt\\-\int_0^x(2+t^3)\,dt\end{gathered}$$. Apply the fundamental theorem and chain rule to both terms:

$$\begin{gathered}A^{\prime}(x)=2\{2+(2x)^3\}\\-(2+x^3)\\=2+15x^3.\end{gathered}$$

The lower endpoint moves too. Its accumulated contribution is subtracted, which produces the negative term.

Illustrative mark guide

• (a) polynomial antiderivative 3; initial condition 1
• (b) upper endpoint including factor 2: 2; negative lower-endpoint term 2; simplification 1; explanation 1

Check: Direct integration gives $A(x)=2x+15x^4/4$; its derivative agrees.

D2: If one endpoint term is missing, split the integral at the fixed point 0. Apply the chain rule separately to each moving endpoint.

Review a guided moving-endpoint calculation

3. A definite integral with an exponential (8 marks)

Let $u=1+e^x$, so $du=e^x\,dx$. At $x=0$, $u=2$; at $x=\ln2$, $u=3$. Thus

$$\begin{gathered}I=\int_2^3u^{-1/2}\,du\\=[2\sqrt u]_2^3\\=2(\sqrt3-\sqrt2).\end{gathered}$$

The transformed variable is positive throughout, so the square root and primitive are valid on the whole interval.

Illustrative mark guide

• substitution and differential 2
• both transformed endpoints 2
• primitive 2
• exact endpoint subtraction 1; sign or size check 1

Check: The integrand is positive. Also $1/\sqrt3\le I\le1/\sqrt2$, since the transformed interval has width 1.

D3: If your integral in u still has x-endpoints, change the endpoints or return to x before evaluating. Do not mix the two variables.

Choosing an integration method

4. An integral with two factors (10 marks)

Integration by parts is useful because differentiating $x$ removes the polynomial factor and cosine has a simple primitive. Choose $u=x$, $dv=\cos x\,dx$; then $du=dx$, $v=\sin x$.

$$\begin{gathered}J=[x\sin x]_0^{\pi/2}\\-\int_0^{\pi/2}\sin x\,dx\\=\frac\pi2-[-\cos x]_0^{\pi/2}\\=\frac\pi2-1.\end{gathered}$$

An antiderivative is $x\sin x+\cos x$. At the lower endpoint it equals 1, so its contribution cannot be dropped.

Illustrative mark guide

• method choice and u, dv 2
• correct integration-by-parts equation 3
• primitive and both endpoint contributions 3
• exact result 1; derivative verification 1

Check: $\frac d{dx}(x\sin x+\cos x)=x\cos x$. The answer is positive, consistent with the integrand on the given interval.

D4: If you obtained π/2, evaluate the whole primitive at 0. If your remaining integral became harder, reconsider which factor you differentiate.

Choosing an integration method

5. Even trigonometric powers (10 marks)

The double-angle and power-reduction identities give $$\begin{gathered}\sin^2x\cos^2x\\=\tfrac14\sin^2(2x)\\=\tfrac18(1-\cos4x)\end{gathered}$$. Therefore

$$\begin{gathered}\int\sin^2x\cos^2x\,dx\\=\frac x8-\frac{\sin4x}{32}+C.\end{gathered}$$

Differentiating gives $1/8-(4/32)\cos4x=(1-\cos4x)/8$, the original integrand by the same identity.

Illustrative mark guide

• valid identities, including both factors 4
• integration with correct frequency factor 3; constant 1
• derivative verification 2

Check: The derivative $(1-\cos4x)/8$ is nonnegative, as $\sin^2x\cos^2x$ must be.

D5: If the cosine coefficient is wrong, reduce the powers in two separate lines. If the antiderivative coefficient is wrong, differentiate the sine term and retain its factor 4.

Choosing an integration method

6. An algebraic radical (12 marks)

Choose $x=2\tan\theta$, with $0\le\theta\le\alpha=\arctan(1/2)<\pi/2$. Then $dx=2\sec^2\theta\,d\theta$. Since $\sec\theta>0$ on this interval, $\sqrt{x^2+4}=2\sec\theta$.

$$\begin{gathered}K=\int_0^\alpha\frac{4\tan^2\theta\,2\sec^2\theta}{8\sec^3\theta}\,d\theta\\=\int_0^\alpha(\sec\theta-\cos\theta)\,d\theta.\end{gathered}$$

Here $$\begin{gathered}\tan^2\theta/\sec\theta\\=(\sec^2\theta-1)/\sec\theta\end{gathered}$$. A primitive is $\ln(\sec\theta+\tan\theta)-\sin\theta$; the logarithm argument is positive on the chosen interval.

At $\alpha$, $\tan\alpha=1/2$, $\sec\alpha=\sqrt5/2$, and $\sin\alpha=1/\sqrt5$. At 0 the primitive is 0. Thus

$$K=\ln\frac{\sqrt5+1}{2}-\frac1{\sqrt5}.$$

Illustrative mark guide

• substitution, differential, angle bounds and square-root sign 3
• correct transformed integral and identity 4
• primitive 3
• exact endpoint values and subtraction 2

Check: In the original variable a primitive is $$\begin{gathered}\ln(x+\sqrt{x^2+4})\\-x/\sqrt{x^2+4}\end{gathered}$$. Its derivative is $$\begin{gathered}1/\sqrt{x^2+4}\\-4/(x^2+4)^{3/2}\\=x^2/(x^2+4)^{3/2}\end{gathered}$$. The numerical value is about 0.0340, which is positive.

D6: If the radical simplification has an unexplained sign, choose an angle interval first. If the factors of 2 do not cancel, transform numerator, denominator and differential separately.

Choosing an integration method

7. A rational integrand (14 marks)

The fraction is proper: the numerator has lower degree than the denominator. The quadratic $x^2+1$ has no real linear factors, so its numerator must be allowed to have degree 1:

$$\begin{gathered}\frac{3x^2+2x+3}{(x+1)(x^2+1)}\\=\frac A{x+1}+\frac{Bx+D}{x^2+1}.\end{gathered}$$

Multiplication by the denominator gives $$\begin{gathered}3x^2+2x+3\\=A(x^2+1)\\+(Bx+D)(x+1)\end{gathered}$$. Comparing coefficients gives $A+B=3$, $B+D=2$, $A+D=3$. Hence $A=2$, $B=1$, $D=1$.

Now integrate $$\begin{gathered}2/(x+1)+x/(x^2+1)\\+1/(x^2+1)\end{gathered}$$. The result is

$$\begin{gathered}2\ln|x+1|+\frac12\ln(x^2+1)\\+\arctan x+C.\end{gathered}$$

The expression is valid on each interval contained in $(-\infty,-1)$ or $(-1,\infty)$. The constant may differ between these disconnected intervals. The second logarithm needs no absolute value because $x^2+1>0$.

Illustrative mark guide

• correct proper-fraction form with linear numerator 3
• coefficient equations and values 4
• three primitive terms 4; arbitrary constant 1
• domain and logarithm conditions 2

Check: Differentiating gives $2/(x+1)+(x+1)/(x^2+1)$. Recombining yields numerator $$\begin{gathered}2(x^2+1)+(x+1)^2\\=3x^2+2x+3\end{gathered}$$.

D7: If your coefficient equations are inconsistent, check whether you omitted the x term over the irreducible quadratic. If you obtained only logarithms, separate x/(x²+1) from 1/(x²+1).

Choosing an integration method

8. Two enclosed regions (12 marks)

Intersections satisfy $x^3=x$, or $x(x-1)(x+1)=0$. Thus $x=-1,0,1$. On $(-1,0)$, $x^3>x$; on $(0,1)$, $x>x^3$.

The total geometric area is

$$\begin{gathered}\mathcal A=\int_{-1}^{0}(x^3-x)\,dx\\+\int_0^1(x-x^3)\,dx.\end{gathered}$$

The first integral is $[x^4/4-x^2/2]_{-1}^0=1/4$. The second is $[x^2/2-x^4/4]_0^1=1/4$. Hence $\mathcal A=1/2$ square coordinate units.

The student’s integral is a signed difference. Its negative and positive contributions cancel because $x-x^3$ is odd. For geometric area, use $\int_{-1}^1|x-x^3|\,dx$, or split where the order of the curves changes.

Illustrative mark guide

• all three intersections 3
• correct upper curves and explanation of signed cancellation 3
• two correct nonnegative integrals and evaluations 4
• combined area and square units 2

Check: Both lobes have positive area and are related by a half-turn about the origin, so their areas must agree.

D8: If an enclosed area is zero or negative, test the order of the curves separately between each pair of intersections. A signed integral can cancel; geometric area cannot.

Volumes and cross-sections

9. A solid about a shifted axis (14 marks)

The rotation axis is horizontal, so use vertical slices of thickness $dx$. On $[1,e]$, $0\le\ln x\le1<2$: the entire region lies below the axis. The lower edge $y=0$ is farther from it.

The outer radius is $R(x)=2$, and the inner radius is $r(x)=2-\ln x$. A typical section is a washer, with a hole of radius $r(x)>0$. Consequently

$$\begin{gathered}V=\pi\int_1^e\{4-(2-\ln x)^2\}\,dx\\=\pi\int_1^e\{4\ln x-(\ln x)^2\}\,dx.\end{gathered}$$

Integration by parts gives $$\begin{gathered}\int_1^e\ln x\,dx\\=[x\ln x-x]_1^e=1\end{gathered}$$. For the square, take $u=(\ln x)^2$, $dv=dx$, so $du=2\ln x\,dx/x$:

$$\begin{gathered}\int_1^e(\ln x)^2dx\\=[x(\ln x)^2]_1^e\\-2\int_1^e\ln x\,dx\\=e-2.\end{gathered}$$

Hence $V=\pi\{4-(e-2)\}=\pi(6-e)$ cubic coordinate units.

The proposed expression squares the strip height $R-r=\ln x$. A washer’s area is $\pi(R^2-r^2)$, not $\pi(R-r)^2$. The proposed disks instead describe rotation of this region about $y=0$.

Illustrative mark guide

• orientation, bounds and both radii 4
• washer model and correct integral 4
• evaluation of both logarithmic integrals and exact result 4
• cubic units and diagnosis of the proposed disk integral 2

Check: Writing $L=\ln x$, with $0\le L\le1$, gives $3L\le4L-L^2\le4L$. Since $\int_1^eL\,dx=1$, the volume must lie between $3\pi$ and $4\pi$. The result $\pi(6-e)$ does.

D9: If the setup is wrong, label the axis and measure both distances from it before integrating. If only the final value is wrong, check the lower endpoint in the logarithmic primitives.

Volumes and cross-sections

Diagnose and retry

1. For $v(t)=2-t$ on $[0,3]$, with the same units as Question 1, write the right-endpoint sum with $n$ equal intervals and find its limit. Does the limit represent distance or net displacement?
2. For $B(x)=\int_{x^2}^{3x}\cos t\,dt$, find $B^{\prime}(x)$ for real $x$.
3. Evaluate $\int_0^{\ln3}\frac{e^x}{2+e^x}\,dx$.
4. Evaluate $\int_1^2\frac{\ln x}{x^2}\,dx$.
5. Find $\int\cos^2(3x)\,dx$.
6. Evaluate $\int_0^1\frac{dx}{(x^2+9)^{3/2}}$ using a trigonometric substitution.
7. Find $\int\frac{x^2+1}{x^2+4}\,dx$. Simplify the rational function first and explain the primitive formula used for the remaining fraction.
8. Find the area between $y=x^2$ and $y=1$ for $0\le x\le2$. Compare it with $\int_0^2(1-x^2)\,dx$.
9. Rotate the region $0\le x\le1$, $0\le y\le x$ about $y=3$. Find its volume with washers.