OMPT drill — Calculus

Integration basics

Antiderivatives and definite integrals as area. Reverse the power rule, mind the +c, and check by differentiating, a ten-second habit that catches most slips. Say the rule you are using out loud before you differentiate or integrate — it sounds silly and it works.

Tested inOMPT-B OMPT-D

Lesson

Integration runs differentiation backwards: an antiderivative of $f$ is any function whose derivative is $f$ . Since $x^{3}$ differentiates to $3 x^{2}$ , an antiderivative of $3 x^{2}$ is $x^{3}$ , and so is $x^{3} + 5$ , and $x^{3} - 12$ , because constants vanish under differentiation. That is why every indefinite integral ends in $+ c$ : you are reporting a whole family of functions, and omitting the $c$ is reporting only one member of it.

The mechanical rule is the power rule in reverse: raise the exponent by one, divide by the new exponent, so $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c$ . The second face of the topic is the definite integral $\int_{a}^{b} f (x) d x$ , which measures the signed area between the graph and the $x$ -axis, and the bridge between the two faces is the cleanest theorem you will use all exam: find any antiderivative $F$ , and the area is simply $F (b) - F (a)$ .

Problem 1Numeric answer

Evaluate $\int_{0}^{2} (3 x^{2} - 2 x) d x$ .

Show the worked solution

Answer: 4

An antiderivative is $x^{3} - x^{2}$ . Evaluating between the bounds: $(8 - 4) - (0 - 0) = 4$ .

Problem 2Multiple choice

Which of the following is an antiderivative of $cos x$ ?

A $- sin x$
B $sin x$
C $- cos x$
D $tan x$

Show the worked solution

Answer: B — $sin x$

I ask: whose derivative is $cos x$ ? Differentiating $sin x$ gives exactly $cos x$ , so $sin x$ works. The minus signs live on the other pairing: $\int sin x d x = - cos x + C$ .

Problem 3Numeric answer

Evaluate $\int_{1}^{4} \frac{1}{x} d x$ .

Show the worked solution

Answer: 2

I rewrite the integrand as $x^{- 1/2}$ ; its antiderivative is $\frac{x ^{1/2}}{1/2} = 2 x$ . Then $24 - 21 = 4 - 2 = 2$ .

Problem 4Multiple choice

What is $\int e^{3 x} d x$ ?

A $e^{3 x} + C$
B $3 e^{3 x} + C$
C $\frac{1}{3} e^{3 x} + C$
D $\frac{e ^{3 x + 1}}{3 x + 1} + C$

Show the worked solution

Answer: C — $\frac{1}{3} e^{3 x} + C$

Differentiating $e^{3 x}$ produces an extra factor 3 by the chain rule, so integrating must compensate with $\frac{1}{3}$ : the answer is $\frac{1}{3} e^{3 x} + C$ . Check by differentiating: $\frac{1}{3} \cdot 3 e^{3 x} = e^{3 x}$ .

Problem 5Spot the error

A student computes $\int_{0}^{π} sin x d x$ . Step 1: an antiderivative of $sin x$ is $cos x$ . Step 2: evaluate at the bounds: $cos π - cos 0$ . Step 3: this equals $- 1 - 1 = - 2$ . Which step contains the error?

Show the worked solution

Answer: Step 1

The antiderivative of $sin x$ is $- cos x$ , not $cos x$ — differentiating $cos x$ gives $- sin x$ . With the sign fixed, the integral is $(- cos π) - (- cos 0) = 1 + 1 = 2$ . Steps 2 and 3 evaluate faithfully; the wrong seed in Step 1 flipped the final sign. An area under a curve that lies above the axis can never be negative, which is the giveaway.

Problem 6Numeric answer

Find the area enclosed between $y = x^{2}$ and $y = 2 x$ . Round to two decimals.

Show the worked solution

Answer: 1.33

The curves meet where $x^{2} = 2 x$ , at $x = 0$ and $x = 2$ , and between them the line lies above the parabola (test $x = 1$ : $2 > 1$ ). So the area is $\int_{0}^{2} (2 x - x^{2}) d x = [x^{2} - \frac{x ^{3}}{3}]_{0}^{2} = 4 - \frac{8}{3} = \frac{4}{3} \approx 1.33$ .