Quick fire โ pick the correct statement in each question. These cover the five Week 6 techniques and the new idea of improper integrals.
$\displaystyle\int e^{5x} \, \text{d}x = \frac{1}{5}e^{5x} + C$ โ reverse chain rule: divide by $a = 5$.
$\displaystyle\int \sec^2(3x) \, \text{d}x = \frac{1}{3}\tan(3x) + C$, since $\dfrac{\text{d}}{\text{d}x}\!\left(\tan(ax+b)\right) = a\sec^2(ax+b)$.
$\sin^2 x = \dfrac{1 - \cos 2x}{2}$, so $\displaystyle\int \sin^2 x \, \text{d}x = \frac{x}{2} - \frac{\sin 2x}{4} + C$.
This is the $\dfrac{f'(x)}{f(x)}$ pattern with $f(x) = x^2+1$, $f'(x) = 2x$. So $\displaystyle\int \frac{6x}{x^2+1} \, \text{d}x = 3\ln|x^2+1| + C$ (factor out 3: $6x = 3 \cdot 2x$).
Use a limit: $\displaystyle\lim_{b \to \infty}\int_{1}^{b} x^{-2}\,\text{d}x = \lim_{b \to \infty}\left[-\frac{1}{x}\right]_{1}^{b} = \lim_{b \to \infty}\left(1 - \frac{1}{b}\right) = 1$. The region is infinite, but the area is finite.
A student sees $\displaystyle\int_{1}^{\infty} \frac{1}{x^2}\,\text{d}x$ and concludes: "The region runs forever, so the integral must diverge."
What went wrong? An infinite region can have a finite area. The integral is defined by a limit: $\displaystyle\lim_{b \to \infty}\int_{1}^{b} \frac{1}{x^2}\,\text{d}x = \lim_{b \to \infty}\left(1 - \frac{1}{b}\right) = 1$. The curve $\dfrac{1}{x^2}$ decays fast enough for the tail to contribute almost nothing. By contrast, $\displaystyle\int_{1}^{\infty} \frac{1}{x}\,\text{d}x = \lim_{b \to \infty}\ln b = \infty$ diverges. Convergence depends on the rate of decay, not on whether the region is unbounded.
The chain rule says $\dfrac{\text{d}}{\text{d}x}\!\left[f(ax+b)\right] = a\,f'(ax+b)$. To reverse, divide by $a$:
Common error: forgetting the $\frac{1}{a}$ factor, or missing the minus sign on $\int \sin(ax+b)$.
Since $\dfrac{\text{d}}{\text{d}x}\!\left(\tan(ax+b)\right) = a\sec^2(ax+b)$, reversing gives:
This is the reverse of the $\tan$ derivative โ remember $\dfrac{\text{d}}{\text{d}x}(\tan x) = \sec^2 x$.
To integrate $\sin^2 x$ or $\cos^2 x$, use the double-angle identities to convert to a form involving $\cos 2x$:
Result: $\displaystyle\int \sin^2 x \, \text{d}x = \frac{x}{2} - \frac{\sin 2x}{4} + C$, โ $\displaystyle\int \cos^2 x \, \text{d}x = \frac{x}{2} + \frac{\sin 2x}{4} + C$
Trap: confusing which identity has $+$ and which has $-$. Remember: $\cos^2 x$ keeps the $+$ sign.
When the numerator's degree is $\geq$ the denominator's degree, the fraction is improper. You must divide first, then decompose the remainder:
1. Polynomial long division: $\dfrac{N(x)}{D(x)} = Q(x) + \dfrac{R(x)}{D(x)}$
2. Decompose $\dfrac{R(x)}{D(x)}$ into partial fractions (now proper).
3. Integrate $Q(x)$ directly and each partial fraction separately.
Example: $\dfrac{4x^2+8x}{2x+3} = 2x + 1 - \dfrac{3}{2x+3}$ (after division), then integrate term by term.
When the numerator is (or is proportional to) the derivative of the denominator, the integral is logarithmic:
If there is a constant multiplier $k$: $\displaystyle\int \frac{k\,f'(x)}{f(x)} \, \text{d}x = k\ln|f(x)| + C$
Example: $\displaystyle\int \frac{6x}{x^2+1} \, \text{d}x = 3\ln(x^2+1) + C$ (here $f(x)=x^2+1$, $f'(x)=2x$, $k=3$).
For a continuous function $f$ with antiderivative $F$:
Three properties to keep straight:
- Reversing the limits changes the sign: $\displaystyle\int_b^a f(x)\,\text{d}x = -\int_a^b f(x)\,\text{d}x$.
- Equal limits give zero: $\displaystyle\int_a^a f(x)\,\text{d}x = 0$.
- The integral is a signed area: regions below the $x$-axis count as negative.
Example: $\displaystyle\int_1^3 (2x+1)\,\text{d}x = \left[x^2+x\right]_1^3 = 12 - 2 = 10$.
An integral is improper when the interval is unbounded or the integrand becomes infinite at an endpoint. In both cases, replace the problem limit with a variable and take a limit.
If the limit is a finite number, the integral converges to that number; if the limit is infinite or does not exist, it diverges.
Two standard examples:
- $\displaystyle\int_1^\infty \frac{1}{x^2}\,\text{d}x = \lim_{b \to \infty}\left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty}\left(1-\frac{1}{b}\right) = 1$ โ converges.
- $\displaystyle\int_1^\infty \frac{1}{x}\,\text{d}x = \lim_{b \to \infty}\left[\ln x\right]_1^b = \lim_{b \to \infty}\ln b = \infty$ โ diverges.
Pattern: $\displaystyle\int_1^\infty \frac{1}{x^p}\,\text{d}x$ converges exactly when $p > 1$.
Trap: never substitute $\infty$ directly into an antiderivative. Write the limit every time โ $\infty$ is not a number.
Split into three experts. Each expert certifies one piece of the definite-integral toolkit โ evaluation, a symmetry property, or an area โ and cannot see what the other two are missing. Only when all three pieces are certified does the combined area problem unlock below, and that problem needs every piece at once.
Reporting rule: when your piece is certified, explain it to the other two in one sentence before the next piece is attempted.
Work on your own. Link every definite integral on the left to its exact value on the right: click an integral, then click the value you predict โ a numbered line joins them. Before you link, predict which value is $0$ and which one is largest. When all six are linked, press Check to score; press Reset to shuffle and play again.
Inquiry tip: reversing the limits changes the sign; an odd integrand over a symmetric interval gives $0$; $\displaystyle\int \dfrac{1}{x}\,\text{d}x$ gives a logarithm, not a power.
Evaluate $\displaystyle\int_{2}^{\infty} \frac{1}{(x-1)^2}\,\text{d}x$, explaining why the integral is improper and whether it converges.
๐ก Hint
Replace the upper limit $\infty$ with $b$, integrate from $2$ to $b$, then take $\lim_{b \to \infty}$.
Use the power rule with $u = x-1$: $\displaystyle\int (x-1)^{-2}\,\text{d}x = -(x-1)^{-1}$.
๐ Reveal Solution
The interval $[2, \infty)$ is unbounded, so the integral is improper. It must be defined by a limit:
$$\int_{2}^{\infty} \frac{1}{(x-1)^2}\,\text{d}x = \lim_{b \to \infty}\int_{2}^{b} (x-1)^{-2}\,\text{d}x.$$
$$\int_{2}^{b} (x-1)^{-2}\,\text{d}x = \left[-\frac{1}{x-1}\right]_{2}^{b} = -\frac{1}{b-1} - \left(-\frac{1}{1}\right) = 1 - \frac{1}{b-1}.$$
$$\lim_{b \to \infty}\left(1 - \frac{1}{b-1}\right) = 1 - 0 = 1.$$
$\boxed{\text{The integral converges to } 1.}$
The integrand is positive on $[2, \infty)$, so the integral must be positive. As $b \to \infty$, $\frac{1}{b-1} \to 0$, leaving $1$. โ
Evaluate $\displaystyle\int_{0}^{1} \frac{1}{x^{1/3}}\,\text{d}x$, explaining why the integral is improper and whether it converges.
๐ก Hint
The integrand tends to $\infty$ as $x \to 0^{+}$, so replace the lower limit $0$ with $t$ and take $\lim_{t \to 0^{+}}$.
Rewrite $\dfrac{1}{x^{1/3}}$ as $x^{-1/3}$ and use the power rule.
๐ Reveal Solution
As $x \to 0^{+}$, $\dfrac{1}{x^{1/3}} \to \infty$, so the integrand is unbounded at the lower endpoint. The integral is improper:
$$\int_{0}^{1} x^{-1/3}\,\text{d}x = \lim_{t \to 0^{+}}\int_{t}^{1} x^{-1/3}\,\text{d}x.$$
$$\int_{t}^{1} x^{-1/3}\,\text{d}x = \left[\frac{3}{2}x^{2/3}\right]_{t}^{1} = \frac{3}{2}\left(1 - t^{2/3}\right).$$
$$\lim_{t \to 0^{+}}\frac{3}{2}\left(1 - t^{2/3}\right) = \frac{3}{2}(1 - 0) = \frac{3}{2}.$$
$\boxed{\text{The integral converges to } \dfrac{3}{2}.}$
The integrand is positive on $(0, 1]$, so the area must be positive. As $t \to 0^{+}$, $t^{2/3} \to 0$, leaving $\frac{3}{2}$. โ
P2 ยง2.5 ยท not examinable in P3
The trapezium rule is a Paper 2 ยง2.5 objective. Candidates taking the P1 + P3 + P4 + P5 route are not examined on it. It is kept here only as enrichment; it is not assessed in the Practice quiz or Exit Ticket.
Error analysis: concave down ($f'' < 0$) โ chords lie below the curve โ underestimate; concave up ($f'' > 0$) โ chords lie above the curve โ overestimate.
๐งฉ Extension example (optional): $\displaystyle\int_1^4 \frac{\ln x}{x}\,\text{d}x$ with 3 strips
$h = \frac{4-1}{3} = 1$, so $x = 1,2,3,4$ and $y = 0,\;0.34657\ldots,\;0.36620\ldots,\;0.34657\ldots$.
$T_3 = \frac{1}{2}\left[0 + 2(0.34657 + 0.36620) + 0.34657\right] \approx 0.89$.
The exact value is $\frac{1}{2}(\ln 4)^2 \approx 0.96$, so the estimate is below the true value. Indeed $y'' = \frac{2\ln x - 3}{x^3} < 0$ on $[1,4]$, so the curve is concave down and the trapezium rule gives an underestimate.
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Evaluate $\displaystyle\int_{1}^{\infty} \frac{1}{x^3}\,\text{d}x$, stating whether the integral converges or diverges.
๐ก Hint
Replace $\infty$ with $b$ and take $\lim_{b \to \infty}$.
Use the power rule: $\displaystyle\int x^{-3}\,\text{d}x = -\frac{1}{2x^2}$.
๐ Reveal Solution
$$\int_{1}^{\infty} \frac{1}{x^3}\,\text{d}x = \lim_{b \to \infty}\int_{1}^{b} x^{-3}\,\text{d}x.$$
$$\int_{1}^{b} x^{-3}\,\text{d}x = \left[-\frac{1}{2x^2}\right]_{1}^{b} = -\frac{1}{2b^2} + \frac{1}{2}.$$
$$\lim_{b \to \infty}\left(\frac{1}{2} - \frac{1}{2b^2}\right) = \frac{1}{2}.$$
$\boxed{\text{The integral converges to } \dfrac{1}{2}.}$
The integrand is positive, so the answer must be positive. As $b \to \infty$, $\frac{1}{2b^2} \to 0$, leaving $\frac{1}{2}$. โ
Evaluate $\displaystyle\int_{0}^{2} \frac{1}{\sqrt{2-x}}\,\text{d}x$, stating whether the integral converges or diverges.
๐ก Hint
The integrand tends to $\infty$ as $x \to 2^{-}$, so replace the upper limit with $t$ and take $\lim_{t \to 2^{-}}$.
Use the power rule directly: $\displaystyle\int (2-x)^{-1/2}\,\text{d}x = -2(2-x)^{1/2}$.
๐ Reveal Solution
As $x \to 2^{-}$, $\dfrac{1}{\sqrt{2-x}} \to \infty$. So
$$\int_{0}^{2} \frac{1}{\sqrt{2-x}}\,\text{d}x = \lim_{t \to 2^{-}}\int_{0}^{t} (2-x)^{-1/2}\,\text{d}x.$$
$$\int_{0}^{t} (2-x)^{-1/2}\,\text{d}x = \left[-2\sqrt{2-x}\right]_{0}^{t} = -2\sqrt{2-t} + 2\sqrt{2}.$$
$$\lim_{t \to 2^{-}}\left(2\sqrt{2} - 2\sqrt{2-t}\right) = 2\sqrt{2} - 0 = 2\sqrt{2}.$$
$\boxed{\text{The integral converges to } 2\sqrt{2}.}$
$2\sqrt{2} \approx 2.828$, positive and finite, as expected for a positive integrand. โ
A curve is such that $\dfrac{\text{d}y}{\text{d}x} = e^{2x} - 2e^{-x}$. The point $(0, 1)$ lies on the curve.
(i) Find the equation of the curve.
(ii) The curve has one stationary point. Find the $x$-coordinate of this point and determine whether it is a maximum or a minimum point.
(9709 Paper 2 Q6 November 2005)
๐ก Hint
For (i): integrate each term. $\int e^{2x} \, \text{d}x = \frac{1}{2}e^{2x}$ (divide by $a = 2$). $\int 2e^{-x} \, \text{d}x = -2e^{-x}$ (divide by $a = -1$, get $-2$). Use $(0, 1)$ to find $C$.
For (ii): set $\frac{\text{d}y}{\text{d}x} = 0$. This gives $e^{2x} = 2e^{-x}$, so $e^{3x} = 2$. Use the second derivative test to classify.
๐ Reveal Solution
Integrate: $y = \displaystyle\int \left(e^{2x} - 2e^{-x}\right) \text{d}x = \frac{1}{2}e^{2x} + 2e^{-x} + C$.
(Note: $\int 2e^{-x} \, \text{d}x = 2 \cdot \frac{1}{-1}e^{-x} = -2e^{-x}$, so subtracting gives $+2e^{-x}$.)
Use $(0, 1)$: $1 = \frac{1}{2}e^{0} + 2e^{0} + C = \frac{1}{2} + 2 + C = \frac{5}{2} + C$.
So $C = 1 - \frac{5}{2} = -\frac{3}{2}$.
$\boxed{y = \frac{1}{2}e^{2x} + 2e^{-x} - \frac{3}{2}}$
Set $\frac{\text{d}y}{\text{d}x} = 0$: $e^{2x} - 2e^{-x} = 0 \implies e^{2x} = 2e^{-x}$.
Multiply by $e^{x}$: $e^{3x} = 2 \implies 3x = \ln 2 \implies x = \frac{1}{3}\ln 2 \approx 0.231$.
Second derivative: $\frac{\text{d}^{2}y}{\text{d}x^{2}} = 2e^{2x} + 2e^{-x}$.
Since $e^{2x} > 0$ and $e^{-x} > 0$ for all $x$, we have $\frac{\text{d}^{2}y}{\text{d}x^{2}} > 0$ always.
$\boxed{x = \frac{1}{3}\ln 2, \text{ and it is a minimum point.}}$
Find the exact value of the constant $k$ for which $\displaystyle\int_{1}^{k} \frac{1}{2x - 1} \, \text{d}x = 1$.
(9709 Paper 3 Q1 November 2007)
๐ก Hint 1
Find the antiderivative: $\int \frac{1}{2x-1} \, \text{d}x = \frac{1}{2}\ln|2x-1|$ (divide by $a = 2$).
Then evaluate at the limits $1$ and $k$ and set the result equal to $1$.
๐ก Hint 2
You should get: $\frac{1}{2}\left(\ln|2k-1| - \ln|2(1)-1|\right) = 1$.
Simplify: $\ln|2(1)-1| = \ln 1 = 0$.
So $\frac{1}{2}\ln(2k-1) = 1$ (assuming $k > \frac{1}{2}$ so $2k-1 > 0$).
Solve for $k$ by exponentiating both sides.
๐ Reveal Solution
Here $a = 2$, $b = -1$. So $\displaystyle\int \frac{1}{2x-1} \, \text{d}x = \frac{1}{2}\ln|2x-1|$.
$\displaystyle\int_{1}^{k} \frac{1}{2x-1} \, \text{d}x = \left[\frac{1}{2}\ln|2x-1|\right]_{1}^{k} = \frac{1}{2}\left(\ln|2k-1| - \ln|2(1)-1|\right)$
$= \frac{1}{2}\left(\ln|2k-1| - \ln 1\right) = \frac{1}{2}\ln|2k-1|$
$\frac{1}{2}\ln|2k-1| = 1 \implies \ln|2k-1| = 2 \implies |2k-1| = e^{2}$
Since the integral is on $[1, k]$ and $2(1)-1 = 1 > 0$, we need $k \geq 1$ so $2k - 1 > 0$.
$2k - 1 = e^{2} \implies k = \dfrac{e^{2} + 1}{2}$
$\boxed{k = \dfrac{e^{2} + 1}{2}}$
Verification: $\frac{1}{2}\ln\!\left(2 \cdot \frac{e^{2}+1}{2} - 1\right) = \frac{1}{2}\ln(e^{2}) = \frac{1}{2} \cdot 2 = 1$. โ
Exit Ticket
Before you leave, tick the objectives you feel confident about. Be honest โ this guides next week's review.