1.8 - Definite integrals

1.8 - Definite integrals

An indefinite integral is a family of functions; a definite integral has limits and gives a value. You will evaluate definite integrals exactly, handle simple finite and infinite endpoints with limits, and use a given integral to find an unknown boundary.

Upper minus lower

Can an integral equal zero even when the integrand is not the zero function? Compare 2x+12x+1 with 2x2x over −1≤x≤1-1\le x\le1. Antiderivatives let us calculate their definite integrals without adding up many small contributions individually.

For a continuous function, a definite integral can be understood as the limit of sums of thin strips: each contribution is the function value multiplied by a small change in the variable. Negative values contribute negatively. The connection with reverse differentiation is that the integral measures the change in an antiderivative.

Evaluating a definite integral

If F′(x)=f(x)F'(x)=f(x) throughout [a,b][a,b], with ff continuous there, then

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

The lower limit is aa; the upper limit is bb. Reversing them changes the sign: ∫baf(x) dx=−∫abf(x) dx\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx.

Not stated in the formula booklet: learn how to apply the limits.

The constant cancels: [F(b)+C]−[F(a)+C]=F(b)−F(a)[F(b)+C]-[F(a)+C]=F(b)-F(a). You may therefore choose one antiderivative and omit +C+C during a definite calculation. Write both substituted expressions so that subtraction acts on the entire lower-limit value.

Worked example 1: retaining a constant term in the integrand

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

Plan. Find one antiderivative and subtract its lower-limit value from its upper-limit value.

∫−11(2x+1) dx=[x2+x]−11=(1+1)−[1+(−1)]=2.\begin{aligned} \int_{-1}^{1}(2x+1)\,dx &=[x^2+x]_{-1}^{1}\\ &=(1+1)-[1+(-1)]\\ &=2. \end{aligned}

Answer. 22.

Check. The linear function's average value over these symmetric limits is 11; multiplying by the interval width 22 gives 22.

Remove only the constant term from the integrand. The contributions on opposite sides of zero now cancel.

Worked example 2: a zero integral

Question. Evaluate ∫−112x dx\int_{-1}^{1}2x\,dx.

Plan. Use the antiderivative x2x^2 at the same two limits.

∫−112x dx=[x2]−11=1−1=0.\int_{-1}^{1}2x\,dx=[x^2]_{-1}^{1}=1-1=0.

Answer. 00, although 2x2x is nonzero at most inputs in the interval.

Check. At opposite inputs −t-t and tt, the function values are −2t-2t and 2t2t. Equal widths therefore contribute equal magnitudes with opposite signs.

A definite integral need not be positive. For instance, ∫13(−2x) dx=[−x2]13=−9−(−1)=−8\int_1^3(-2x)\,dx=[-x^2]_1^3=-9-(-1)=-8. Reversing these limits gives 88. Exact fractional or surd answers should remain exact unless a decimal is requested.

Watch out

[x2+x]−11[x^2+x]_{-1}^1 means (12+1)−[(−1)2+(−1)](1^2+1)-[(-1)^2+(-1)]. Do not subtract only the first term, and do not reverse upper minus lower.

Keep brackets around each substituted value in this new calculation.

Improper endpoint integrals

The function 1/x1/\sqrt{x} is undefined at x=0x=0, so the ordinary continuous-interval rule cannot be applied there directly. Yet the accumulated integral towards that endpoint can approach a finite value. A tall, narrow contribution need not make the total infinite.

The same question arises with an infinite endpoint: the interval is unbounded, but its contributions may become small quickly enough. Infinity is not a number to substitute into an antiderivative; it describes a limiting process.

Improper integral and convergence

For a problematic lower endpoint, first use a permitted finite boundary ε>0\varepsilon>0, then let ε\varepsilon approach 00 from above:

∫0bf(x) dx=lim⁡ε→0+∫εbf(x) dx.\int_0^b f(x)\,dx=\lim_{\varepsilon\to0^+}\int_\varepsilon^b f(x)\,dx.

For an infinite upper endpoint, use a finite boundary RR, then increase it without bound:

∫a∞f(x) dx=lim⁡R→∞∫aRf(x) dx.\int_a^\infty f(x)\,dx=\lim_{R\to\infty}\int_a^R f(x)\,dx.

An improper integral converges if the limit is finite; otherwise it diverges.

The symbols ε→0+\varepsilon\to0^+ mean positive values approaching zero. We only need elementary power behaviour: ε→0\sqrt{\varepsilon}\to0, while 1/ε1/\sqrt{\varepsilon} increases without bound. Use this after integrating on the valid truncated interval.

Worked example 3: an undefined endpoint with a finite integral

Question. Evaluate ∫04x−1/2 dx\int_0^4 x^{-1/2}\,dx.

Plan. Integrate from ε>0\varepsilon>0 to 44, then examine the lower-end limit.

∫ε4x−1/2 dx=[2x]ε4=4−2ε.\begin{aligned} \int_\varepsilon^4 x^{-1/2}\,dx &=[2\sqrt{x}]_\varepsilon^4\\ &=4-2\sqrt{\varepsilon}. \end{aligned}

As ε→0+\varepsilon\to0^+, the subtracted term tends to zero.

Answer. The integral converges to 44.

Check. Lower boundaries 0.010.01 and 0.00010.0001 give 3.83.8 and 3.983.98, approaching 44 from below. The function remains undefined at zero.

Change only the exponent to −3/2-3/2. The antiderivative now contains a reciprocal root, which has different endpoint behaviour.

Worked example 4: an endpoint that makes the integral diverge

Question. Determine whether ∫04x−3/2 dx\int_0^4 x^{-3/2}\,dx converges.

Plan. Use the same truncation and inspect the resulting expression.

∫ε4x−3/2 dx=[−2x−1/2]ε4=−1+2ε.\int_\varepsilon^4 x^{-3/2}\,dx =[-2x^{-1/2}]_\varepsilon^4 =-1+\frac2{\sqrt{\varepsilon}}.

The last term increases without bound as ε→0+\varepsilon\to0^+.

Answer. The integral diverges; it has no finite value.

Check. The same lower boundaries 0.010.01 and 0.00010.0001 give 1919 and 199199. Their increase agrees with the limiting expression.

For an infinite endpoint, use powers of RR instead. For example, 1/R→01/R\to0 as R→∞R\to\infty, whereas R\sqrt{R} grows without bound.

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