Integration Workbook: 10 Exercises with Full Solutions

This workbook accompanies the Integration study guide and is built for practice with pen and paper. The ten exercises move through the same arc as the lesson: the power rule and standard integrals, the linear combination rule, definite integrals and area, then the three big techniques of substitution, integration by parts, and partial fractions. Each one isolates a single skill so you can pinpoint exactly where your method needs work.

Attempt every exercise before reading the solutions, and write out each step rather than jumping to the answer. The fastest check on any integral is to differentiate your result, so several solutions end by doing exactly that. If your derivative returns the original integrand, your integration was correct.

Part One: The Exercises

Exercise 1 (Power rule and linear combination). Find the indefinite integral of 3x squared plus 4x minus 5.

Exercise 2 (Negative and fractional powers). Find the indefinite integral of two over x plus the square root of x.

Exercise 3 (Exponential and trigonometric). Find the indefinite integral of e to the x plus 3 cos x.

Exercise 4 (Definite integral). Evaluate the definite integral of 2x plus 1 from x equals 1 to x equals 3.

Exercise 5 (Area, non-negative integrand). Find the area between the curve y equals x squared, the x-axis, and the lines x equals 0 and x equals 2.

Exercise 6 (Area below the axis). Find the area between the curve y equals x squared minus 4, the x-axis, and the lines x equals 0 and x equals 2.

Exercise 7 (Substitution, linear inner function). Find the indefinite integral of (2x plus 1) to the power 4.

Exercise 8 (Substitution, non-linear inner function). Find the indefinite integral of x times e to the x squared.

Exercise 9 (Integration by parts). Find the indefinite integral of x times e to the x.

Exercise 10 (Partial fractions, distinct linear factors). Find the indefinite integral of 1 over (x minus 1)(x plus 2).

Part Two: Worked Solutions

Solution 1. Split with the linear combination rule and apply the power rule to each term, raising each power by one and dividing.

Differentiating gives 3x squared plus 4x minus 5, confirming the result.

Solution 2. Rewrite both terms as powers of x first: two over x is 2 times x to the minus one, and the square root is x to the one half.

The first term uses the special reciprocal rule giving a logarithm, and the second uses the ordinary power rule. The absolute value in the logarithm is needed because x may be negative.

Solution 3. Each term is a standard integral, so apply them directly through the linear combination rule.

Note that the integral of cosine is positive sine; it is the integral of sine that carries the minus sign.

Solution 4. Integrate to find the antiderivative, then evaluate it at the upper limit minus the lower limit, omitting the constant since it cancels.

Solution 5. On the interval from 0 to 2 the curve y equals x squared is non-negative, so the area equals the definite integral directly.

The area is eight thirds.

Solution 6. Here the integrand needs a check first. On the interval from 0 to 2, x squared minus 4 is negative, since x squared stays below 4, so the curve lies below the axis and the integral will come out negative.

The area is the magnitude of this value, sixteen thirds. Reporting minus sixteen thirds as the area would be the classic mistake.

Solution 7. The inner function is linear, so substitute g equal to 2x plus 1, which gives dx equal to one half dg.

The factor of one half appears because the derivative of the inner function is 2.

Solution 8. The inner function x squared is non-linear, so substitution only works because the integrand also contains an x, which matches its derivative. Set g equal to x squared, so dg equals 2x dx and x dx equals one half dg.

Differentiating with the chain rule returns x times e to the x squared, as required.

Solution 9. This is a product of unlike functions, so use integration by parts. The polynomial x simplifies when differentiated, so let f be x and g prime be e to the x, giving f prime equal to 1 and g equal to e to the x.

The new integral was simpler than the original, which is the whole point of the method.

Solution 10. The integrand is a rational function with distinct linear factors, so decompose it into partial fractions.

Clearing denominators gives 1 equal to A times (x plus 2) plus B times (x minus 1). Setting x equal to 1 gives A equal to one third, and setting x equal to minus 2 gives B equal to minus one third. Each piece then integrates to a logarithm.

How to Get the Most From This Workbook

Notice the decision that sits in front of every one of these problems: before you integrate, you identify which kind of integrand you are facing. A standard form or a linear combination integrates directly. A composition with a linear inner function, or a product where one factor is the derivative of the other’s inner part, calls for substitution. A product of unlike functions calls for parts, and a rational function calls for partial fractions. Drill that recognition step until it is automatic, keep the area rule’s sign trap in mind by always checking whether the curve dips below the axis, and verify by differentiating whenever you are unsure. Do that, and integration stops feeling like guesswork and becomes a matter of matching the integrand to the right tool.

See you soon.