OMPT drill — Calculus

Integration techniques

Substitution carries the load at OMPT level. Recognising the inner function and its derivative sitting next to each other is the entire game. 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

The reverse power rule only reaches so far. Try it on $\int 2 x (x^{2} + 1)^{3} d x$ and it has nothing to say, because the integrand is not a plain power of $x$ : it is a composite, a function of a function, with something extra multiplied on. Substitution is the tool for exactly this shape, and at OMPT level it carries essentially the whole load of this topic. It is the chain rule run backwards, the same way basic antiderivatives were the power rule run backwards.

The pattern to hunt for: an inner function, and its derivative (or a constant multiple of it) sitting next to it as a factor. In $2 x (x^{2} + 1)^{3}$ , the inner function is $x^{2} + 1$ and there is its derivative, $2 x$ , standing right beside it. When you spot that pairing, substitution will work; when the pairing is absent, no amount of clever $u$ -choosing will manufacture it. Recognising the pair is the entire game: the mechanics that follow are four rehearsed lines.

Problem 1Numeric answer

Evaluate $\int_{0}^{1} 2 x (x^{2} + 1)^{3} d x$ . Round to two decimals.

Show the worked solution

Answer: 3.75

The factor $2 x$ is exactly the derivative of $x^{2} + 1$ , so I substitute $u = x^{2} + 1$ , $d u = 2 x d x$ . The bounds become $u = 1$ to $u = 2$ , and the integral is $\int_{1}^{2} u^{3} d u = [\frac{u ^{4}}{4}]_{1}^{2} = \frac{16 - 1}{4} = \frac{15}{4} = 3.75$ .

Problem 2Multiple choice

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

A $\frac{x ^{2}}{2} e^{x} + C$
B $(x - 1) e^{x} + C$
C $(x + 1) e^{x} + C$
D $x e^{x} + C$

Show the worked solution

Answer: B — $(x - 1) e^{x} + C$

Integration by parts with $u = x$ and $d v = e^{x} d x$ gives $x e^{x} - \int e^{x} d x = x e^{x} - e^{x} = (x - 1) e^{x} + C$ . Differentiating the answer confirms it: $e^{x} + (x - 1) e^{x} = x e^{x}$ .

Problem 3Numeric answer

Evaluate $\int_{0}^{π /2} x cos x d x$ . Round to two decimals.

Show the worked solution

Answer: 0.57

By parts with $u = x$ , $d v = cos x d x$ : the antiderivative is $x sin x + cos x$ . Evaluating: at $\frac{π}{2}$ it gives $\frac{π}{2} \cdot 1 + 0 = \frac{π}{2}$ ; at 0 it gives $0 + 1 = 1$ . The integral is $\frac{π}{2} - 1 \approx 0.57$ .

Problem 4Multiple choice

Which substitution best handles $\int \frac{x}{1 - x ^{2}} d x$ ?

A $u = x$
B $u = x$
C $u = 1 - x^{2}$
DIntegration by parts with $u = x$

Show the worked solution

Answer: C — $u = 1 - x^{2}$

With $u = 1 - x^{2}$ , $d u = - 2 x d x$ , the stray $x$ in the numerator is absorbed: the integral becomes $- \frac{1}{2} \int u^{- 1/2} d u = - u + C = - 1 - x^{2} + C$ . The presence of the inner function's derivative (up to a constant) is the signal for this substitution.

Problem 5Spot the error

A student computes $\int ln x d x$ by parts. Step 1: choose $u = ln x$ and $d v = d x$ , so $d u = \frac{1}{x} d x$ and $v = x$ . Step 2: apply the formula as $uv + \int v d u = x ln x + \int x \cdot \frac{1}{x} d x$ . Step 3: simplify to $x ln x + x + C$ . Which step contains the error?

Show the worked solution

Answer: Step 2

The choices in Step 1 are the standard ones and are correct. Step 2 misquotes the formula: integration by parts reads $\int u d v = uv - \int v d u$ , with a minus sign. Correctly, $\int ln x d x = x ln x - \int 1 d x = x ln x - x + C$ . Step 3 simplifies the wrong expression without adding new mistakes.

Problem 6Numeric answer

Evaluate $\int_{1}^{e} \frac{ln x}{x} d x$ . Give a decimal.

Show the worked solution

Answer: 0.5

Setting $u = ln x$ gives $d u = \frac{1}{x} d x$ , and the bounds map to $u = 0$ and $u = 1$ . The integral becomes $\int_{0}^{1} u d u = \frac{1}{2}$ .