CEGEP Calculus II — The Definite Integral and the Fundamental Theorem Worksheet

All 7 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Area Under a Curve and Riemann Sums

   Let $f(x)=x^2+1$ on $[0,2]$.

   1. Estimate the area under the graph of $f$ over $[0,2]$ using the right-endpoint Riemann sum $R_4$ (four subintervals of equal width).
   2. The exact area is $\frac{14}{3}$. State whether $R_4$ over- or under-estimates it, and justify your answer from the shape of the graph rather than from the two numbers.
2. Q2The Definite Integral as a Limit of a Riemann Sum

   Evaluate ${\displaystyle {\int }_0^3(x^2+1)dx}$ directly from the definition, as the limit of right-endpoint Riemann sums. You may use ${\displaystyle {\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}}$ and ${\displaystyle {\sum }_{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}}$.

3. Q3Properties of the Definite Integral

   Suppose $f$ and $g$ are continuous and $${\int }_1^{5}f(x)dx=8,{\int }_5^{9}f(x)dx=6,{\int }_1^{5}g(x)dx=-3.$$

   1. Find ${\displaystyle {\int }_1^5(2f(x)-3g(x))dx}$.
   2. Find ${\displaystyle {\int }_9^{1}f(x)dx}$.
   3. Explain why the data above is not enough to determine ${\displaystyle {\int }_1^{5}f(x)g(x)dx}$, and support the explanation with an example.
4. Q4The Fundamental Theorem, Part One

   Use the Fundamental Theorem of Calculus, Part One, to differentiate each accumulation function. Do not attempt to evaluate the integrals.

   1. $F(x)={\displaystyle {\int }_3^x\frac{t}{1+t^4}dt}$. Give $F^{\prime }(x)$, and state the value of $F(3)$.
   2. $G(x)={\displaystyle {\int }_x^5{e}^{-t^2}dt}$.
   3. $H(x)={\displaystyle {\int }_0^{x^3}\sqrt{1+t^2}dt}$.
5. Q5The Fundamental Theorem, Part Two

   Evaluate each definite integral with an antiderivative, showing the antiderivative used.

   1. ${\displaystyle {\int }_1^4(3\sqrt{x}-\frac{2}{x^2})dx}$
   2. ${\displaystyle {\int }_1^e\frac{(\ln x{)}^2}{x}dx}$
6. Q6Average Value and the Mean Value Theorem for Integrals

   Over one twelve-hour daytime period the temperature inside a greenhouse is modelled by $$T(t)=18+6\sin \left(\frac{\pi t}{12}\right)\text{degrees Celsius},0\le t\le 12,$$ where $t$ is the number of hours after opening. A calculator is needed for the decimals requested below.

   1. Find the average temperature over the twelve hours, exactly and to three decimal places.
   2. The Mean Value Theorem for Integrals guarantees at least one time $c$ at which the temperature equals that average. State why its hypothesis holds here, then find every such $c$ in $[0,12]$, to three decimal places.
7. Q7Synthesis — drawing on several topics in this set

   Let ${\displaystyle I={\int }_0^4(x^2+2)dx}$.

   1. Write $I$ as a limit of right-endpoint Riemann sums and evaluate that limit, using ${\displaystyle {\sum }_{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}}$.
   2. Evaluate $I$ again with the Fundamental Theorem, Part Two, and confirm the two agree.
   3. Using the properties of the definite integral — not a fresh computation — state the values of ${\displaystyle {\int }_4^0(x^2+2)dx}$ and ${\displaystyle {\int }_0^4(3x^2+6)dx}$.

The 5 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

An estimate needs a reason, not a comparison

Q1 asks for a right-endpoint sum R₄ and then, deliberately, tells you the exact value before asking whether the estimate runs high or low. The point of handing you the exact value is that you may not use it: part (b) asks you to justify the direction from the shape of the graph. "One number came out bigger than the other" is an observation, not a justification, and it earns nothing.

Two habits go with this. Write Δx = (b − a) ⁄ n down explicitly and list the endpoints before summing anything — a left sum and a right sum over the same interval share all but one term each, and reaching for the wrong list is the commonest arithmetic slip on the sheet. And say what your number is: an area in square units, or, when the integrand is a rate, an amount.

Evaluating from the definition, in a fixed order

Q2 wants a definite integral computed as the limit of right-endpoint Riemann sums, with the summation formulas supplied. It is a long calculation with no decisions in it, which is exactly why it is worth drilling: every step is forced, and the marks are for carrying n through the whole thing rather than for the answer.

Q7 then asks for the same integral twice — once from the definition, once with the Fundamental Theorem — and the point is not that the two agree. It is that they must, and that the theorem replaces an infinite process by a two-point subtraction. That is the whole content of the course to this point, and doing both routes on one integral is the only way most students actually feel it.

The properties are structural: they never need the function

Q3 gives you three numerical values and asks for others, without ever naming f or g. You evaluate by rearranging, not by integrating.

Part One and Part Two are two different theorems

They share a name, and students blur them constantly. This set keeps them on separate concepts because they take different inputs, produce different outputs and fail for different reasons. Q4 is Part One; Q5 is Part Two.

Part One differentiates. It takes a function with a variable limit — an accumulation function — and hands back the integrand, with the dummy variable replaced by the upper limit. No antiderivative is ever found, which is the point: it applies perfectly well to integrands like t ⁄ (1 + t⁴) or e^−t², whose antiderivatives you cannot write down in elementary terms at all. Q4 is deliberately built from such integrands, and it tells you not to evaluate the integrals so that you cannot dodge into Part Two.

Part Two evaluates. It takes a definite integral with two numbers for limits and returns a number, and it does need an antiderivative. Part One's output is a function; Part Two's output is a value.

Two variants appear in Q4, and both are ordinary Part One once the shape is repaired first.

Part Two has a hypothesis, and it is not decoration

Q5 is two evaluations, and both ask you to show the antiderivative you used. Part (b), with (ln x)² ⁄ x, is a substitution, and there are two clean routes: convert the limits to u as you substitute and finish entirely in u, or keep the original limits, convert back to x and only then evaluate. Both are correct. Mixing them — evaluating a function of u at limits that were written for x — gives a wrong answer whose working looks flawless.

So before writing brackets, look at the interval and the integrand together: does anything in a denominator vanish inside it, does a logarithm meet a non-positive argument, does an even root meet a negative one. Then evaluate — and remember that a definite integral never carries + C, because the constant cancels in F(b) − F(a). Every antiderivative on this sheet can be checked in one line by differentiating it back.

Average value: the height of the rectangle with the same area

Q6 models the temperature inside a greenhouse and asks for its average over twelve hours. The average value of a function is not the average of a few readings — it weights every instant, and it does so through the integral.

Read it as the height of the rectangle over [a, b] enclosing the same area as the region under the graph. The 1 ⁄ (b − a) is the factor that gets dropped, and dropping it turns an average temperature into an accumulated temperature-hour — which the units expose at once, if you write them.

Getting the most out of it

Name the property, not the direction

Wherever the sheet asks why, the mark is on the reason. "The right sum overestimates because the function is increasing, so the right endpoint is where it is largest on each subinterval" is a complete answer; "the right sum overestimates" is not, even when it is correct. The same holds for the Mean Value Theorem for Integrals: state that the function is continuous on the closed interval, then use the conclusion.

Before Part Two, look at the interval

Make it automatic. Read the integrand, read the limits, and ask whether the integrand is continuous everywhere between them. It costs three seconds, and it is the difference between an answer and a number.

Differentiate your antiderivative before moving on

Every antiderivative on this sheet can be verified in one line, and the check catches the usual slips: a missing chain-rule factor, a lost constant multiple, a sign. Differentiating is the operation you are already fluent in — use it on your own work.

Say what the number means, with units

An integral of a rate is an amount; an average value carries the units of the function, not of the integral. Translating the number into a sentence is a mark of its own on most college finals, and it is the fastest error check you have — an interpretation that sounds absurd usually is.

Common questions