Flashcards · 12 cards

Integration

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Rule
Integrate $ax^n$ .
Answer
$\dfrac{ax^{n + 1}}{n + 1} + c$ for $n \ne -1$ : add 1 to the power, divide by the new power, add $c$ .
∫ a xⁿ dx = a x^{n + 1} ⁄ (n + 1) + c
add 1 to the power, then divide by the new power; never forget + c (n ≠ −1)
Integrating a powerAdd 1 to the power, divide by the new power, add c
From the lesson: Integration techniques
Rule
Integration by substitution: the steps?
Answer
Let $u$ be the inside; find $du = u'\,dx$ ; swap everything, $dx$ included, into $u$ ; integrate; put $x$ back.
∫ 2x(x² + 1)³ dx: let u = x² + 1, so du = 2x dx
= ∫ u³ du = u⁴ ⁄ 4 + c = (x² + 1)⁴ ⁄ 4 + c
let u be the inside; its derivative should appear outside; change the limits too
Substitutionu = the inside; du = u′ dx; swap everything into u
From the lesson: Integration techniques
Rule
$\displaystyle\int \frac{f'(x)}{f(x)}\,dx = \,?$
Answer
$\ln|f(x)| + c$ , when the top is the derivative of the bottom.
∫ 1 ⁄ (ax + b) dx = (1 ⁄ a) ln|ax + b| + c
∫ f′(x) ⁄ f(x) dx = ln|f(x)| + c
when the top is the derivative of the bottom, the answer is a log
Integrals that give logs∫ f′(x) ÷ f(x) dx = ln|f(x)| + c
From the lesson: Integration techniques
Know it
Integrating a fraction: what must you not do?
Answer
Integrate the top and bottom separately. Simplify it, split it into partial fractions, or substitute first.
From the lesson: Integration techniques
Rule
Part of the curve is below the $x$ -axis. How do you find the total area?
Answer
Integrate each part separately: the part below comes out negative. Add the sizes.
Above and below the axisThe integral counts the part below the axis as negative
From the lesson: Areas, volumes and motion
Rule
The volume of revolution about the $y$ -axis?
Answer
$V = \pi\int_a^b x^2\,dy$ , where $a$ and $b$ are $y$ -values.
About the y-axisV = π∫ x² dy: discs of radius x, limits a and b are y-values
From the lesson: Areas, volumes and motion
Rule
From acceleration to velocity and displacement?
Answer
Integrate: $v = \int a\,dt$ and $s = \int v\,dt$ , using the starting values to find each constant.
distance sd/dt →velocity vd/dt →acceleration a
differentiate to go right; integrate to come back
Displacement, velocity, accelerationDifferentiate down the chain; integrate back up it
From the lesson: Areas, volumes and motion
Know it
The velocity changes sign during the time asked about. How do you find the distance travelled?
Answer
Split the integral where $v = 0$ (the particle turns back there) and add the sizes.
From the lesson: Areas, volumes and motion
Rule
The trapezium rule?
Answer
$\int_a^b y\,dx \approx \dfrac h2\big[y_0 + y_n + 2(y_1 + \dots + y_{n - 1})\big]$ : the first and last ordinates once, the middle ones twice.
Strips as trapeziumsEach strip is about a trapezium: ½h(yᵢ + yᵢ₊₁)
From the lesson: The trapezium rule
Know it
Five ordinates from $a$ to $b$ : how many strips, and what is $h$ ?
Answer
Four strips, so $h = \dfrac{b - a}{4}$ .
From the lesson: The trapezium rule
Which method?
WAEC 2018 · Paper 2 · Q11 (b)
Find $\displaystyle\int x\sqrt{1 - x}\,dx$ .
Which substitution?
Answer
Let $u = 1 - x$ , so $x = 1 - u$ and $dx = -du$ . The integral becomes $-\int (1 - u)u^{\frac12}\,du$ , which splits into powers of $u$ .
From the lesson: Integration techniques Try the question
Which method?
WAEC 2023 · Paper 2 · Q2
Using the trapezium rule with five ordinates, evaluate, correct to two decimal places, $\displaystyle\int_0^2 \frac{4}{1 + x^2}\,dx$ .
Approximate value
What is $h$ ?
Answer
Five ordinates make four strips, so $h = \dfrac{2 - 0}{4} = 0.5$ , with ordinates at $x = 0, 0.5, 1, 1.5, 2$ .
From the lesson: The trapezium rule Try the question