Module 16: Antiderivatives, Riemann sums, and definite integrals

8 exercises, each with a complete solution. 1 figure.

On this page

A. Prerequisites and observable goals
B. The central idea: add many small contributions
C. Definitions before calculation shortcuts
D. A sum that becomes an exact area
E. How to recognize the requested object
F. Worked teaching examples
G. Independent practice
H. Complete solutions: Module 16
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Derivative rules, finite sums, limits, and signed rates.

After

Distinguish a family of antiderivatives from an integral value, construct a Riemann sum, and interpret signed accumulation with units.

B. The central idea: add many small contributions#

Differentiation extracts a local rate from a function. Accumulation goes in the other direction: many small rate-times-width contributions add up to a total. The definite integral makes this idea exact by taking a limit of sums. An antiderivative is a function whose derivative is known; the Fundamental Theorem in the next module explains why these seemingly different objects are connected.

Add many small contributions. Six rectangles of width

0.5

under

y=1+\tfrac12x^2

[0,3]

, each as tall as the curve at its left edge. The highlighted one contributes

f(1.5)\cdot0.5

. Their total is

6.4375

; the exact area is

7.5

, and more, thinner rectangles close the gap.

C. Definitions before calculation shortcuts#

An antiderivative of $f$ on an interval is a function $F$ with $F'=f$ . Any two antiderivatives on the same interval differ by a constant, by the MVT. The notation $\int f(x)\dd x=F(x)+C$ describes a family of functions, not a single number. On disconnected domains, separate constants may be needed on separate intervals.

For a partition $a=x_0<x_1<\cdots<x_n=b$ , let $\Delta x_i=x_i-x_{i-1}$ and choose $x_i^*\in[x_{i-1},x_i]$ . A Riemann sum is $\sum_{i=1}^n f(x_i^*)\Delta x_i$ . If all such sums converge to one value as the largest subinterval width tends to zero, that value is the definite integral

\int_a^b f(x)\dd x=\lim_{\|P\|\to0}\sum_{i=1}^n f(x_i^*)\Delta x_i.

A continuous function on a closed interval is Riemann integrable. Many discontinuous functions are also integrable, so continuity is sufficient, not necessary. For equal widths, $\Delta x=(b-a)/n$ and right endpoints are $a+i\Delta x$ . Integral foundations: [O1Int].

The integral is signed: portions below the horizontal axis contribute negatively. Geometric area uses $\int|f|$ , splitting at sign changes when necessary. Reversing bounds changes sign, and splitting at an interior input adds accumulations. If $f$ is measured in meters per second and $x$ in seconds, the integral has meters as its units.

D. A sum that becomes an exact area#

For $f(x)=x$ on $[0,2]$ , a right-endpoint sum has width $2/n$ and sample inputs $2i/n$ :

S_n=\sum_{i=1}^n\frac{2i}{n}\frac2n =\frac4{n^2}\frac{n(n+1)}2=2+\frac2n.

Thus $\int_0^2 x\dd x=2$ , agreeing with the area of a triangle of base $2$ and height $2$ . Every finite right sum overestimates the increasing function; the limit removes the $2/n$ discrepancy. The identity $\sum i=n(n+1)/2$ is a finite algebraic fact, while the passage $2+2/n\to2$ is a limit.

E. How to recognize the requested object#

An indefinite integral asks for a family of functions; differentiate your answer. A definite integral asks for an accumulated number; its units and bounds matter. A numerical sum asks for a finite approximation and must not be silently replaced by the limiting answer. To build a sum, identify width, sample input, function value, and number of terms in that order. The sample input is not always just $i/n$ .

F. Worked teaching examples#

Undoing a derivative. For $f(x)=6x^2-4x+5$ , termwise antidifferentiation gives $F(x)=2x^3-2x^2+5x+C$ . Checking gives $F'=6x^2-4x+5$ . If $F(0)=7$ , then $C=7$ ; an initial value selects one member of the family.

Signed displacement. Suppose velocity is $3$ m/s for two seconds and then $-2$ m/s for three seconds. Displacement is $3(2)+(-2)(3)=0$ m. Distance is $3(2)+2(3)=12$ m. Zero net accumulation does not mean nothing happened.

A midpoint setup to finish. For $\int_1^3 x^2\dd x$ with four equal subintervals, the width is $1/2$ . The midpoints are $5/4,7/4,9/4,11/4$ . Thus the midpoint sum is $\tfrac12[(5/4)^2+(7/4)^2+(9/4)^2+(11/4)^2]=69/8$ . The exact integral, obtained next module, is $26/3$ , so the sum is an approximation.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M16.01FSolution

Find all antiderivatives of $8x^3-6x+2$ on $\R$ , and the one with $F(1)=5$ .

M16.02CSolution

Find an antiderivative of $3/x+2\cos x$ on $(0,\infty)$ . Explain why $3\ln x$ rather than $3x^0/0$ is used.

M16.03CSolution

Compute the left and right sums with four equal subintervals for $f(x)=x+1$ on $[0,2]$ . Compare them with the exact geometric integral.

M16.04TSolution

Build a right-endpoint Riemann sum for $\int_2^5\sqrt{x}\dd x$ with $n$ equal intervals. Do not evaluate the limit.

M16.05CSolution

Use $\sum_{i=1}^n i^2=n(n+1)(2n+1)/6$ to evaluate $\int_0^1 x^2\dd x$ directly from right-endpoint sums.

M16.06TSolution

A rate is $4$ liters/min for $0\le t<2$ and $-1$ liter/min for $2\le t\le5$ . Find net change, total variation in volume, and final volume if the initial volume is $10$ liters.

M16.07TSolution

Given $\int_1^4 f=7$ and $\int_1^2 f=-3$ , find $\int_2^4 f$ , $\int_4^1 f$ , and $\int_1^4 2f$ . Are these data enough to find the geometric area under $f$ ?

M16.08SSolution

On $[-1,1]$ , let $g(x)=0$ except $g(0)=5$ . Explain using Riemann sums why $\int_{-1}^1g(x)\dd x=0$ despite the discontinuity.

Optional small hints

For M16.05, separate powers of $n$ before taking the limit. For M16.08, only subintervals containing zero can contribute, and at most two such intervals occur in a partition.

H. Complete solutions: Module 16#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M16.01Show solution

Integrate, then use the initial value

The power rule gives $F=2x^4-3x^2+2x+C$ . At $1$ , $F(1)=2-3+2+C=1+C$ , so $C=4$ . The selected function is $\boxed{2x^4-3x^2+2x+4}$ . Differentiating verifies the integrand; substituting $1$ verifies the initial condition.

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M16.02Show solution

The power rule’s excluded exponent

Because $(\ln x)'=1/x$ for $x>0$ , an antiderivative is $\boxed{3\ln x+2\sin x+C}$ . The usual rule $\int x^p\dd x=x^{p+1}/(p+1)+C$ excludes $p=-1$ , where its denominator is zero. On a negative interval, the logarithmic primitive would be $\ln|x|$ instead.

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M16.03Show solution

Finite sums and their exact target

The width is $1/2$ . Left sample heights are $1,3/2,2,5/2$ , so $L_4=(1/2)(7)=\boxed{7/2}$ . Right heights are $3/2,2,5/2,3$ , so $R_4=(1/2)(9)=\boxed{9/2}$ . The trapezoid under the line has area $2(1+3)/2=\boxed4$ . Increasing behavior explains $L_4<4<R_4$ .

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M16.04Show solution

Retain the interval’s offset

Here $\Delta x=(5-2)/n=3/n$ and the $i$ th right input is $2+3i/n$ . Therefore

\boxed{\int_2^5\sqrt x\dd x=\lim_{n\to\infty}\sum_{i=1}^n\sqrt{2+3i/n}\,\frac3n}.

The starting input $2$ belongs inside the square root. The external $3/n$ is a width, not part of the function input.

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M16.05Show solution

Evaluate an integral from its defining sums

The width is $1/n$ and sample input is $i/n$ . Hence

S_n=\frac1{n^3}\sum_{i=1}^n i^2 =\frac{n(n+1)(2n+1)}{6n^3} =\frac{(1+1/n)(2+1/n)}6\longrightarrow\boxed{1/3}.

Continuity guarantees that this right-sum limit is the integral, not a special result that depends on the chosen sample convention.

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M16.06Show solution

Accumulate the rate over each time interval

Net change is $4(2)-1(3)=\boxed5$ liters. Total variation adds magnitudes, $8+3=\boxed{11}$ liters. Final volume is initial volume plus net change, $10+5=\boxed{15}$ liters. The integral gives a change, not the final amount without the initial value.

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M16.07Show solution

Use linearity and interval additivity

Since $7=-3+\int_2^4f$ , the latter integral is $\boxed{10}$ . Reversing bounds gives $\boxed{-7}$ , and multiplying the integrand by $2$ gives $\boxed{14}$ . The signed data alone do not determine $\int_1^4|f|$ : positive and negative contributions inside the smaller intervals may cancel in many different ways.

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M16.08Show solution

A single exceptional height has no width

Every Riemann-sum term is zero unless its sample input is zero. At most two subintervals can have zero as an endpoint or interior point. If the mesh is $\|P\|$ , their total contribution is nonnegative and at most $10\|P\|$ . As the mesh tends to zero, every Riemann sum is squeezed to zero. Thus the integral is $\boxed0$ . This proves integrability directly and shows why continuity is not a necessary condition.

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I. Retention and readiness#

Revisit M05.05 and M12.01. Explain why an integral of a rate has amount units, why $+C$ belongs to an indefinite answer, and why a point’s height alone supplies no area. Advance when you can construct a sum on an interval that does not begin at zero.