1.08d, 1.08e, 1.08g - Definite integrals, signed area and limits of sums | OCR A-Level Mathematics A

Evaluate definite integrals, distinguish signed accumulation from geometric area and split at changes of sign. Explain why narrow rectangle sums approach an exact integral.

A definite integral is a number

Can an integral be zero while the graph encloses a positive area? For the line $y=x-2$ from $x=0$ to $x=4$ , the two triangular regions each have area $2$ . We will see why their integral is nevertheless zero, and why rectangles can recover both answers.

You already know

If $F'(x)=f(x)$ , then $F$ is an antiderivative. An indefinite integral gives $F(x)+C$ ; a definite integral has specified endpoints.

The endpoints are inputs, not heights. Substitute each into the entire antiderivative.

Evaluating a definite integral

For a function continuous on $[a,b]$ , with $F'(x)=f(x)$ ,

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

The constant cancels: $(F(b)+C)-(F(a)+C)=F(b)-F(a)$ .

Not in the formula sheet: learn it.

A bracketed antiderivative keeps the endpoint substitution visible, particularly when a lower endpoint is negative.

Worked example 1: evaluate both endpoints

Question. Evaluate $\int_0^2(2x^2-3x+1)\,\mathrm dx$ .

Plan. Integrate term by term, then subtract the lower endpoint value.

\begin{aligned} \int_0^2(2x^2-3x+1)\,\mathrm dx &=\left[\frac{2x^3}{3}-\frac{3x^2}{2}+x\right]_0^2 &&\text{antiderivative}\\ &=\left(\frac{16}{3}-6+2\right)-0 &&\text{upper minus lower}\\ &=\frac43 &&\text{simplify exactly} \end{aligned}

Why this step? Each endpoint belongs in every term before you take the difference.

Answer. $4/3$ .

Check. Differentiating the bracket gives $2x^2-3x+1$ .

Change only the lower endpoint. It now contributes a nonzero value.

Worked example 2: a nonzero lower contribution

Question. Evaluate $\int_1^2(2x^2-3x+1)\,\mathrm dx$ .

Plan. Use the same antiderivative and subtract its value at $1$ .

\begin{aligned} \int_1^2(2x^2-3x+1)\,\mathrm dx &=\left[\frac{2x^3}{3}-\frac{3x^2}{2}+x\right]_1^2\\ &=\frac43-\left(\frac23-\frac32+1\right) &&\text{subtract the whole value}\\ &=\frac43-\frac16=\frac76 \end{aligned}

Why this step? A nonzero lower limit cannot be ignored just because an earlier example started at zero.

Answer. $7/6$ .

Check. The omitted integral from $0$ to $1$ is $1/6$ , and $1/6+7/6=4/3$ .

Reversing bounds reverses the sign, equal bounds give zero, and adjoining intervals add: $\int_a^b f=\int_a^c f+\int_c^b f$ . These are properties of signed accumulation, not rules for geometric area.

Watch out

Endpoint subtraction does not justify crossing an undefined point. For example, $1/x^2$ is undefined at zero, so plugging $-1$ and $1$ into an antiderivative does not evaluate an ordinary definite integral across zero.

Now make just the lower endpoint negative.

Signed accumulation and positive area

An integral counts below-axis contributions negatively. A geometric area counts both sides positively.

Signed integral and geometric area

For $a<b$ , $\int_a^b f(x)\,\mathrm dx$ adds signed contributions. If $f\ge0$ , it equals the area enclosed by the curve, the axis and the ordinates $x=a,x=b$ . If $f\le0$ , that area is $-\int_a^b f(x)\,\mathrm dx$ . An ordinate here is a vertical boundary line. Area is non-negative; an integral need not be.

The line from the opening puzzle makes the distinction visible.

Positive area and negative signed contribution

The below-axis triangle contributes $-2$ to the integral, although its area is $2$ .

Worked example 3: unequal regions

Question. Find the signed integral and geometric area for $y=x-2$ on $0\le x\le3$ .

Plan. Split at the root $2$ , evaluate each piece and add their magnitudes for area.

\begin{aligned} \int_0^2(x-2)\,\mathrm dx&=\left[\frac{x^2}{2}-2x\right]_0^2=-2 &&\text{below the axis}\\ \int_2^3(x-2)\,\mathrm dx&=\left[\frac{x^2}{2}-2x\right]_2^3=\frac12 &&\text{above the axis}\\ I&=-2+\frac12=-\frac32,\qquad A=2+\frac12=\frac52 \end{aligned}

Why this step? Sign changes belong in the integral; their magnitudes belong in the geometric area.

Answer. The signed integral is $-3/2$ ; the area is $5/2$ square units.

Check. Triangle areas are $2\times2/2=2$ and $1\times1/2=1/2$ .

Change only the upper boundary, from $3$ to $4$ . The two triangular contributions now balance.

Worked example 4: zero integral, nonzero area

Question. Find the signed integral and geometric area for $y=x-2$ on $0\le x\le4$ .

Plan. Keep the negative contribution and evaluate the enlarged positive piece.

\begin{aligned} \int_0^2(x-2)\,\mathrm dx&=-2\\ \int_2^4(x-2)\,\mathrm dx&=\left[\frac{x^2}{2}-2x\right]_2^4=2\\ I&=-2+2=0,\qquad A=2+2=4 \end{aligned}

Why this step? Equal areas on opposite sides cancel in signed accumulation, not in geometry.

Answer. The integral is $0$ , while the area is $4$ square units.

Check. The two triangles have the same base and height, each $2$ .

If both axes measure metres, these geometric areas are in $\mathrm{m^2}$ ; otherwise use square units for abstract coordinates.

Watch out

Wrong: $A=|\int_0^3(x-2)\,\mathrm dx|=3/2$ . Taking the magnitude after cancellation loses area. Split first, then add $|-2|+|1/2|=5/2$ .

Extend just the upper boundary once more.

Locate every change of sign

For a curved graph, find the roots before choosing the signs of its integral pieces. A quick sign table can save a page of wrong area working.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Mathematics

Statistics

Mechanics

1.08d, 1.08e, 1.08g - Definite integrals, signed area and limits of sums

A definite integral is a number

Signed accumulation and positive area

Locate every change of sign

Continue this lesson.

Learn faster with Nova