9A, 9C - Definite integrals

An indefinite integral gives a family of functions; a definite integral gives a number for a specified interval. You will evaluate it exactly, interpret its sign and work backwards to find a missing limit or parameter, ready for later area and motion applications.

Evaluating between limits

If two antiderivatives can differ by any constant, how can an integral between two fixed numbers have one answer?

For $f(x)=2x$ , choose $F(x)=x^2$ . Its change from $x=1$ to $x=3$ is $9-1=8$ . Choosing $x^2+5$ instead gives $(9+5)-(1+5)=8$ again: the constant cancels.

Evaluating a definite integral

If $F'(x)=f(x)$ , and $f$ is continuous on the interval from $a$ to $b$ ,

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

The number at the bottom is the lower limit and the number at the top is the upper limit. Substitute into the antiderivative, then subtract the whole lower-limit value from the whole upper-limit value.

A definite answer needs no $+C$ . Keep fractions and standard trigonometric values exact; a decimal is an approximation unless it is itself exact.

Worked example 1: upper value minus lower value

Question. Find the exact value of $\int_1^3(2x+1)\,dx$ .

Plan. Integrate first, then evaluate the antiderivative at both limits.

\begin{aligned} \int_1^3(2x+1)\,dx &=[x^2+x]_1^3 &&\text{an antiderivative is }x^2+x\\ &=(3^2+3)-(1^2+1) &&\text{upper minus lower}\\ &=12-2=10 \end{aligned}

Answer. $10$ .

Check. The integrand is a straight line with endpoint values $3$ and $7$ . Its average value over the interval is $(3+7)/2=5$ ; multiplying by the width $2$ gives $10$ , agreeing with the integral.

Keep the integrand and upper limit but move the lower limit to a negative number. Substitute that number with its sign intact.

Worked example 2: a negative lower limit

Question. Find $\int_{-1}^3(2x+1)\,dx$ .

Plan. Use the same antiderivative and enclose the complete lower-limit substitution in brackets.

\begin{aligned} \int_{-1}^3(2x+1)\,dx &=[x^2+x]_{-1}^3\\ &=(9+3)-\big((-1)^2+(-1)\big)\\ &=12-(1-1)=12 \end{aligned}

Answer. $12$ .

Check. The straight-line endpoint values are $-1$ and $7$ . Their average is $3$ , and the interval width is $4$ , giving $3\times4=12$ . Adding an arbitrary constant to the antiderivative would not change the result.

Keep brackets around a negative substituted number before raising it to a power.

Signed accumulation

A definite integral records a signed net contribution, so positive and negative portions can cancel.

With limits running from smaller to larger, a graph above the horizontal axis contributes positively and a graph below it contributes negatively. One way to picture this is to add narrow strips of width $\Delta x>0$ , each with signed contribution approximately $f(x)\Delta x$ . The integral is the exact accumulation as the strips become narrower.

[DIAGRAM: asset_name: Cancellation in a definite integral; asset_slug: signed-contributions; caption: Equal positive and negative contributions give a zero net integral; description: The graph y equals x minus 1 runs from x equals 0 to x equals 2. The portion below the axis from 0 to 1 has contribution negative one half; the portion above from 1 to 2 has contribution positive one half. Their sum is zero, though the graph is not zero throughout the interval.]
Equal positive and negative contributions give a zero net integral

For that line, $\int_0^1(x-1)\,dx=-\frac12$ and $\int_1^2(x-1)\,dx=\frac12$ . Adding contributions across adjacent intervals gives $\int_0^2(x-1)\,dx=0$ . A zero integral therefore does not mean the function was zero everywhere. A geometric area counts sizes positively; a definite integral preserves signs.

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9A, 9C - Definite integrals

Evaluating between limits

Signed accumulation

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