Integration flashcards

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Integration
The reverse process of differentiation: given a gradient function f′(x), find the functions f(x) that it came from.
Indefinite integral
An integral with no limits, ∫f(x) dx = F(x) + c, where F′(x) = f(x). It describes a whole family of functions.
Constant of integration
The arbitrary constant c added to every indefinite integral, because any constant differentiates to zero.
Antiderivative (an integral of f)
Any function F with F′(x) = f(x). Two antiderivatives of the same function differ by a constant.
The dx in ∫f(x) dx
Shows the variable you are integrating with respect to; it goes with the ∫ sign.
Integrand
The expression being integrated — the f(x) in ∫f(x) dx.
∫xⁿ dx
xⁿ⁺¹/(n + 1) + c, for any rational n ≠ −1. (No — learn it)
∫a dx (a constant)
ax + c (the case n = 0). (No — derive it)
∫k f(x) dx and ∫(f(x) ± g(x)) dx
k∫f(x) dx; ∫f(x) dx ± ∫g(x) dx. One + c covers all the terms. (No — learn it)
√x, 1/xⁿ and 1/√x as powers
x^(1/2), x^(−n), x^(−1/2). (No — learn it)
Equation of a curve from f′(x) and a point (x₁, y₁)
f(x) = ∫f′(x) dx; substitute x = x₁, y = y₁ to find c. (No — derive it)
Graphs of y = F(x) + c for different values of c
Vertical translations of one another: at any x they all have the same gradient f(x).
Where f′(x) = 0 on the graph of the gradient function
The x-values where the curve y = f(x) has zero gradient (its stationary points).
Where the graph of f′(x) lies below the x-axis
The curve y = f(x) is decreasing there.
How a given point picks one curve from the family y = F(x) + c
Exactly one member of the family passes through the point; substituting it fixes c.
Shape of y = 2√x, the integral of 1/√x with c = 0
Starts at the origin with a vertical tangent and rises ever more gently, with gradient 1/√x.
Why does every indefinite integral need + c?
Differentiating removes any constant, so infinitely many functions share the same derivative.
Why can the power rule not integrate x⁻¹?
The new power would be n + 1 = 0, so the rule would divide by zero. ∫x⁻¹ dx needs the natural logarithm (P3).
Why must products be expanded before integrating?
There is no rule ∫fg dx = ∫f dx × ∫g dx; differentiating such a product does not give back fg.
How do you check an integral quickly?
Differentiate your answer; you should get back the integrand exactly.
Is the constant of integration always the y-intercept?
No. c = f(0) only when every other term is zero at x = 0; terms like 12/x are not even defined there.
“The gradient is zero at x = a” in an integration question
It is a condition on the gradient: f′(a) = 0. A point on the curve is a condition on f.
∫(8x³ − 3x² + 7) dx
2x⁴ − x³ + 7x + c.
∫6√x dx
6x^(1/2) → 6 × x^(3/2) ÷ 3/2 = 4x^(3/2) + c.
∫4/x² dx
4x⁻² → 4 × x⁻¹ ÷ (−1) = −4/x + c.
∫(x + 4)(x − 2) dx
Expand: x² + 2x − 8. Integral: x³/3 + x² − 8x + c.
∫(√x + 1)/√x dx, x > 0
Split: 1 + x^(−1/2). Integral: x + 2x^(1/2) + c.
f′(x) = 3x² − 2 and the curve passes through (2, 5). Find f(x).
f(x) = x³ − 2x + c; 8 − 4 + c = 5, so c = 1: f(x) = x³ − 2x + 1.
dy/dx = 3√x and the curve passes through (9, 50). Find y.
y = 2x^(3/2) + c; 2 × 27 + c = 50, so c = −4: y = 2x^(3/2) − 4.
Integrating x³ to get 3x²
That is differentiation. Raise the power and divide: x⁴/4 + c.
Integrating x^(−1/2)
New power is +1/2, and dividing by 1/2 doubles: 2x^(1/2) + c.
Integrating the top and bottom of a fraction separately
Never. Split a sum over a single term into separate powers of x first.
Forgetting to find c when a point is given
The final answer must be the full equation, with c replaced by its value.
Substituting the given point into f′(x)
The point lies on the curve, so it goes into f(x).
“Hence” after rewriting an expression
Integrate the rewritten form from the previous part.

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