Mc22, Mc23, Mc25 - Definite integrals, signed area and limits of sums

Mc22, Mc23, Mc25 - Definite integrals, signed area and limits of sums

Evaluate definite integrals, separate signed accumulation from geometric area and interpret an integral as a limit of rectangles. Read a finite-sum argument and explain its bounds, error and exact limit.

A definite integral is a number

Can two positive areas add to a zero integral? On 0≤x≤10\le x\le1, the line y=2x−1y=2x-1 forms two triangles, each of area 1/41/4, but its integral is zero. The missing distinction is sign. We will build the calculation from antiderivatives, then connect it with narrow rectangles.

You already know

An indefinite integral is a family F(x)+cF(x)+c, where F′(x)=f(x)F'(x)=f(x). The same constant disappears when one function value is subtracted from another.

The notation for a definite integral records which inputs to use.

Definite integration

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

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

It gives a number, whereas an indefinite integral gives a family. The same constant in F(b)+cF(b)+c and F(a)+cF(a)+c cancels.

Not in the formula sheet: learn it.

The lower limit may contribute a negative value. Put it in brackets before subtracting.

Worked example 1: a negative lower limit

Question. Evaluate ∫−11(3x2+2) dx\int_{-1}^1(3x^2+2)\,\mathrm dx.

Plan. Integrate the polynomial and evaluate upper minus lower.

∫−11(3x2+2) dx=[x3+2x]−11antiderivative=(1+2)−(−1−2)both endpoint values=6subtract the negative value\begin{aligned} \int_{-1}^1(3x^2+2)\,\mathrm dx&=[x^3+2x]_{-1}^1 &&\text{antiderivative}\\ &=(1+2)-(-1-2) &&\text{both endpoint values}\\ &=6 &&\text{subtract the negative value} \end{aligned}

Why this step? The lower contribution is −3-3, so subtracting it adds 33.

Answer. 66.

Check. The integrand is even, so twice the integral from 00 to 11 also gives 66.

Change only the upper endpoint. The same antiderivative still applies.

Worked example 2: extend the interval

Question. Evaluate ∫−12(3x2+2) dx\int_{-1}^2(3x^2+2)\,\mathrm dx.

Plan. Evaluate the same bracket at 22 and −1-1.

∫−12(3x2+2) dx=[x3+2x]−12=(8+4)−(−1−2)new upper value=15\begin{aligned} \int_{-1}^2(3x^2+2)\,\mathrm dx&=[x^3+2x]_{-1}^2\\ &=(8+4)-(-1-2) &&\text{new upper value}\\ &=15 \end{aligned}

Why this step? Changing a boundary changes the accumulation interval; it does not change the integrand's antiderivative.

Answer. 1515.

Check. The extra integral from 11 to 22 is 12−3=912-3=9, and 6+9=156+9=15.

Reversing the bounds reverses the sign; equal bounds give zero. Splitting at cc gives ∫abf=∫acf+∫cbf\int_a^b f=\int_a^c f+\int_c^b f. Linearity permits sums and constant multiples to be integrated term by term.

Watch out

A bracket calculation across an undefined point is not an ordinary definite integral. For example, 1/x21/x^2 is undefined at zero; endpoint substitution across zero does not fix that problem.

Vary the upper boundary once more and keep the lower bracket intact.

Signed accumulation and positive area

The integral uses signed heights. Geometric area uses non-negative heights, so cancellation must be prevented.

Area between a graph and the axis

For a<ba<b, the geometric area is enclosed by the graph, the axis and the vertical boundaries x=a,x=bx=a,x=b. It equals ∫abf(x) dx\int_a^b f(x)\,\mathrm dx when f≥0f\ge0, and −∫abf(x) dx-\int_a^b f(x)\,\mathrm dx when f≤0f\le0. If the sign changes, split into regions and add their magnitudes.

Here the line crosses at x=1/2x=1/2. The left region's height is negative in the integral, but its area is positive.

Signed contributions of a line crossing at one half

Start with unequal regions, then move just one boundary.

Worked example 3: signed sum and area

Question. For y=2x−1y=2x-1 on [0,2][0,2], find the signed integral and geometric area.

Plan. Split at 1/21/2 and use the antiderivative x2−xx^2-x.

∫01/2(2x−1) dx=[x2−x]01/2=−14∫1/22(2x−1) dx=[x2−x]1/22=94I=−14+94=2,A=14+94=52\begin{aligned} \int_0^{1/2}(2x-1)\,\mathrm dx&=[x^2-x]_0^{1/2}=-\frac14\\ \int_{1/2}^2(2x-1)\,\mathrm dx&=[x^2-x]_{1/2}^2=\frac94\\ I&=-\frac14+\frac94=2,\qquad A=\frac14+\frac94=\frac52 \end{aligned}

Why this step? Only the area calculation reverses the sign of a below-axis contribution.

Answer. Integral 22; area 5/25/2 square units.

Check. The triangle areas are (1/2)×1/2=1/4(1/2)\times1/2=1/4 and (3/2)×3/2=9/4(3/2)\times3/2=9/4.

Change only the upper boundary to 11. Now the two triangles match.

Worked example 4: cancellation

Question. For y=2x−1y=2x-1 on [0,1][0,1], find the signed integral and geometric area.

Plan. Evaluate the two contributions either side of 1/21/2.

I1=[x2−x]01/2=−14I2=[x2−x]1/21=14I=0,A=14+14=12\begin{aligned} I_1&=[x^2-x]_0^{1/2}=-\frac14\\ I_2&=[x^2-x]_{1/2}^1=\frac14\\ I&=0,\qquad A=\frac14+\frac14=\frac12 \end{aligned}

Why this step? A zero net accumulation says the signed contributions balance, not that no region exists.

Answer. Integral 00; area 1/21/2 square unit.

Check. The two triangles have equal bases 1/21/2 and equal heights 11.

Use square units for an abstract coordinate area. If both coordinates measure metres, the area unit is m2\mathrm{m^2}.

Watch out

Taking the magnitude at the end is too late: ∣−1/4+9/4∣=2|-1/4+9/4|=2 is not the area 5/25/2. Take each region's magnitude before adding.

Change just the upper limit again.

Locate every change of sign

Roots divide the interval into regions, but you still need to check each sign. Touching the axis is not the same as crossing it: x2x^2 has a root at zero and remains non-negative.

A quadratic with positive outer regions and a negative central region

For x2−4x^2-4, the roots are −2,2-2,2; the sign is positive outside and negative between them. Use the stated vertical boundaries even when they cut through a lobe.

Worked example 5: three regions

Question. Find the geometric area and signed integral for y=x2−4y=x^2-4 on [−3,3][-3,3].

Plan. Split at −2,2-2,2 and evaluate with F(x)=x3/3−4xF(x)=x^3/3-4x.

I1=[F(x)]−3−2=73above the axisI2=[F(x)]−22=−323below the axisI3=[F(x)]23=73above the axisA=7+32+73=463,I=−6\begin{aligned} I_1&=[F(x)]_{-3}^{-2}=\frac73 &&\text{above the axis}\\ I_2&=[F(x)]_{-2}^{2}=-\frac{32}{3} &&\text{below the axis}\\ I_3&=[F(x)]_2^3=\frac73 &&\text{above the axis}\\ A&=\frac{7+32+7}{3}=\frac{46}{3},\qquad I=-6 \end{aligned}

Why this step? A single endpoint subtraction gives the net integral, but it hides how much area was cancelled.

Answer. Area 46/346/3 square units; signed integral −6-6.

Check. The equal outer contributions agree with the symmetry of the graph.

Change only the upper endpoint to 22. One positive outer region is removed.

Worked example 6: a boundary at a root

Question. Find the geometric area and signed integral for y=x2−4y=x^2-4 on [−3,2][-3,2].

Plan. Keep only the left and central pieces from the previous calculation.

I=73−323=−253signed contributionsA=73+323=13magnitudes\begin{aligned} I&=\frac73-\frac{32}{3}=-\frac{25}{3} &&\text{signed contributions}\\ A&=\frac73+\frac{32}{3}=13 &&\text{magnitudes} \end{aligned}

Why this step? An area beyond the supplied boundary cannot be included merely because it is part of a familiar sketch.

Answer. Area 1313 square units; signed integral −25/3-25/3.

Check. F(2)−F(−3)=−16/3−3=−25/3F(2)-F(-3)=-16/3-3=-25/3.

The entirely below-axis region on [−2,2][-2,2] has area 32/332/3, the magnitude of its integral. An area is positive even when there is no positive part of the graph.

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