P2.8 - Integration

Learning objectives

P2.8.1Evaluation of definite integralsEvaluation of definite integrals.P2.8.2Interpretation of the definiteInterpretation of the definite Students will be expected to be able to evaluate the area of integral as the area under a curve. a region bounded by a curve and given straight lines.; For example, find the finite area bounded by the curve y = 6x − x2 and the line y = 2x. ∫ x dy will not be required.; Students will be expected to be able to evaluate the area of a region bounded by two curves.P2.8.3Trapezium rule for area approximationApproximation of area under a For example, curve using the trapezium rule. use the trapezium rule to approximate ∫ (2x +1) dx using four strips.; Use of increasing number of trapezia to improve accuracy and an estimate of the error may be required.

A definite integral gives the accumulated signed value of a function between two bounds. First find an antiderivative $F$ with $F'(x)=f(x)$ , then evaluate upper bound minus lower bound.

$\int_a^b f(x)\,dx=[F(x)]_a^b=F(b)-F(a)$

For example, $\int_1^3(2x^2-4x+5)\,dx=\left[\frac23x^3-2x^2+5x\right]_1^3=15-\frac{11}{3}=\frac{34}{3}.$ Write the substitution line before simplifying; this makes the order of subtraction visible.

Reversing the bounds changes the sign: $\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$ . Adjacent intervals add, so $\int_a^c f=\int_a^b f+\int_b^c f$ ; this can reveal a missing integral without integrating again.

Do not subtract lower minus upper, and do not leave $+C$ in a definite answer. Any constants of integration would cancel in $F(b)-F(a)$ .

A definite integral is signed, but geometric area is non-negative. Sketch or compare the boundaries first, find their intersections, and integrate the vertical height of the region with respect to x.

$A=\int_a^b(\text{upper }y-\text{ lower }y)\,dx$

The curves $y=6x-x^2$ and $y=2x$ meet where $6x-x^2=2x$ , so $x=0$ or $x=4$ . Between these values the quadratic is above the line. Hence $A=\int_0^4\big[(6x-x^2)-2x\big]dx=\left[2x^2-\frac{x^3}{3}\right]_0^4=\frac{32}{3}.$

Region	Integrand
curve above the $x$ -axis	$y$
curve below the $x$ -axis	$-y$
between $y=f(x)$ and $y=g(x)$	upper function $-$ lower function

If the upper boundary changes, or a curve crosses the x-axis inside the region, split the integral at that x-value and make each piece positive. Straight-line boundaries are handled by the same upper-minus-lower rule.

Solving for intersections supplies the limits; it does not supply the area. Do not integrate lower minus upper and then report a negative geometric area. This objective uses vertical strips and $dx$ ; $\int x\,dy$ is outside the required scope.

The trapezium rule replaces a curve by straight chords over equal-width strips. With n strips from a to b, calculate the width and list all n+1 ordinates before applying the endpoint weights.

$h=\frac{b-a}{n},\qquad T_n=\frac h2\left[y_0+y_n+2(y_1+\cdots+y_{n-1})\right]$

For $\int_0^1\sqrt{2x+1}\,dx$ with four strips:

$x$	0	0.25	0.50	0.75	1.00
$y$	1.0000	1.2247	1.4142	1.5811	1.7321

Here $h=0.25$ , so $T_4=\frac{0.25}{2}\{1.0000+1.7321+2(1.2247+1.4142+1.5811)\}\approx1.3965.$ Keep extra calculator digits until the final rounding.

More strips make h smaller and usually improve the approximation. If a trusted exact or more accurate value is available, the estimated absolute error is the absolute difference from the trapezium estimate. The chord picture also explains direction: chords below the curve give an underestimate; chords above it give an overestimate.

There are $n+1$ ordinates for $n$ strips. Count each endpoint once and each interior ordinate twice; do not use unequal x-spacing in this formula, and do not round table values too early.

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—Pearson Edexcel IAL MathematicsRead

P2.8 - Integration

Learning objectives

Evaluate a definite integral from its bounds

Turn a bounded region into a non-negative integral

Approximate an integral with equal-width trapezia

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