9709 · 1.8
Integration flashcards
Revision flashcards for Cambridge 9709 Integration (syllabus 1.8). Flip, recall, then mark a real past-paper question.
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What is integration?
The process of finding a function (the integral) whose derivative is a given function. It is the reverse process of differentiation.
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What is the power rule for integrating $x^n$?
$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, for any real number $n \neq -1$. The rule is: 'add one to the power, then divide by the new power'.
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What is an indefinite integral?
An integral without limits, e.g., $\int f(x) \, dx$. The result is a family of functions, not a single value, represented by the constant of integration, $+C$.
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What is a definite integral?
An integral with upper and lower limits, e.g., $\int_a^b f(x) \, dx$. The result is a single numerical value representing the net area under the curve from $x=a$ to $x=b$.
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Why is the '+ C' necessary for indefinite integrals?
The derivative of any constant is zero. When we integrate, we don't know if there was an original constant term, so we add '+ C' to represent all possible constant values.
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Why don't we write '+ C' for definite integrals?
When evaluating $[F(x)+C]_a^b$, the calculation becomes $(F(b)+C) - (F(a)+C) = F(b) - F(a)$. The constant $C$ always cancels out.
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How do you integrate a term like $\frac{5}{\sqrt{x}}$?
First, rewrite it in index form: $5x^{-1/2}$. Then apply the power rule: $\int 5x^{-1/2} \, dx = 5 \times \frac{x^{1/2}}{1/2} + C = 10x^{1/2} + C = 10\sqrt{x} + C$.
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What happens if you integrate to find an area that is below the x-axis?
The definite integral will yield a negative value. The actual area is the positive magnitude of this result. For example, if $\int_1^3 f(x) \, dx = -5$, the area is 5.
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What's the first step to integrating an expression like $(x-3)(2x+1)$?
Expand the brackets first. $(x-3)(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$. Then integrate this polynomial term by term.
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How do you find the equation of a curve if you only know its gradient function $\frac{dy}{dx}$?
Integrate the gradient function to get the general equation $y = F(x) + C$. You then need a point $(x,y)$ that lies on the curve to substitute in and solve for the specific value of $C$.