SACE Methods definite integrals: worked examples

Practise SACE Stage 2 Methods definite integrals using exact values, nonzero bounds and signed area. These are original revision questions, with worked reasoning and a short free trial.

Evaluate an antiderivative at both bounds and subtract the entire lower-bound value.
Interpret the sign of a definite integral, including intervals where the function lies below the horizontal axis.
Find zeros, check the sign on each interval and split an area calculation wherever the sign changes.

The idea

The fundamental theorem of calculus connects differentiation and integration. For a continuous function on a closed interval, evaluate an antiderivative at the upper bound and subtract its value at the lower bound. A definite integral gives signed area; regions below the horizontal axis contribute negatively.

Before you begin: Apply the power rule for antidifferentiation and substitute into a polynomial.

Worked examples

01 Start with the method

Evaluate the definite integral exactly.

\int_{0}^{2} (3 x^{2} + 1) d x

Only the region from 0 to 2 is shaded. The function is positive here, so the integral equals its area: 10 square units.

Step 1

An antiderivative is found by increasing each power by one and dividing by the new power. Differentiate it to check.

F (x) = x^{3} + x,

F^{'} (x) = 3 x^{2} + 1

Step 2

Evaluate at the upper bound and subtract the lower-bound value.

\int_{0}^{2} (3 x^{2} + 1) d x = {[x^{3} + x]}_{0}^{2}

Step 3

Substitute both bounds, including the zero. Here the function is positive throughout the interval, so the integral also equals the area under the curve.

(2^{3} + 2) - (0^{3} + 0) = 10

A mistake to avoid

Subtract the whole lower-bound value. A constant of integration cancels in the subtraction, so it is not part of a definite integral’s numerical answer.

02A nonzero lower bound: subtract a whole expression

Evaluate the definite integral exactly. Keep the lower-bound substitution in brackets until you have simplified it.

\int_{1}^{3} (3 x^{2} - 2 x + 2) d x

Step 1

Antidifferentiate each term. Differentiate the result mentally to check that every coefficient returns to its original value.

F (x) = x^{3} - x^{2} + 2 x,

F^{'} (x) = 3 x^{2} - 2 x + 2

Step 2

The upper bound is three and the lower bound is one. Evaluate the same antiderivative at both; the bounds are not new powers.

\int_{1}^{3} (3 x^{2} - 2 x + 2) d x

= F (3) - F (1)

Step 3

Subtract the whole lower-bound value. Evaluate each bracket before subtracting so that none of its terms is missed.

F (3) = 27 - 9 + 6 = 24

F (1) = 1 - 1 + 2 = 2

24 - 2 = 22

Step 4

This integrand is positive throughout the interval, so its signed integral also equals the total area between the curve and the horizontal axis here. Completing the square shows why it is positive.

3 x^{2} - 2 x + 2

= 3 {(x - \frac{1}{3})}^{2} + \frac{5}{3}

> 0

Result

\int_{1}^{3} (3 x^{2} - 2 x + 2) d x

= 22

03A crossing: signed integral and total area are different

For the given function and interval, calculate both the definite integral and the total area between the curve and the horizontal axis.

f (x) = x - 2,

0 \leq x \leq 5

Step 1

Find where the graph meets the horizontal axis, then check the sign on either side. The function is negative before the crossing and positive after it.

x - 2 = 0 \Rightarrow x = 2

f (x) < 0 (0 \leq x < 2)

f (x) > 0 (2 < x \leq 5)

Step 2

For the signed integral, use the bounds directly. The part below the axis contributes negatively and cancels some of the part above.

\int_{0}^{5} (x - 2) d x

= {[\frac{x^{2}}{2} - 2 x]}_{0}^{5}

= \frac{25}{2} - 10

= \frac{5}{2}

Step 3

For total area, split at the zero where the sign changes. Calculate each signed contribution before turning the negative contribution into a positive area.

\int_{0}^{2} (x - 2) d x = - 2,

\int_{2}^{5} (x - 2) d x = \frac{9}{2}

Step 4

Add the magnitudes of the two regions. Taking the absolute value of the original integral would give five halves, not the total area of thirteen halves.

A = - \int_{0}^{2} (x - 2) d x

+ \int_{2}^{5} (x - 2) d x

= 2 + \frac{9}{2}

= \frac{13}{2}

Result

Signed integral = \frac{5}{2}

A = \frac{13}{2} square units

Practice questions with answers

Work these out first, then open each solution to compare your method. These questions are untimed.

Question 1

Evaluate with a nonzero lower bound.

\int_{1}^{3} 2 x d x

Show solution 1

{[x^{2}]}_{1}^{3} = 9 - 1 = 8

Using only the upper-bound value would give 9 and miss the subtraction at the lower bound.

Question 2

Evaluate exactly without a calculator.

\int_{2}^{4} (2 x + 1) d x

Show solution 2

{[x^{2} + x]}_{2}^{4}

= (16 + 4) - (4 + 2)

= 20 - 6 = 14

An antiderivative is x squared plus x. Its upper-bound value is 20 and its lower-bound value is 6; subtract both complete values.

Question 3

Evaluate exactly. Explain why the result is negative even though the integrand is positive between one and three.

\int_{3}^{1} 2 x d x

Show solution 3

{[x^{2}]}_{3}^{1} = 1 - 9

= - 8

The upper bound is one, even though it is the smaller number. Reversing the bounds reverses the sign; this is a signed integral, not a request for total area.

Question 4

Find the signed integral and the total area between this curve and the horizontal axis over the stated interval.

f (x) = x,

- 1 \leq x \leq 2

Show solution 4

\int_{- 1}^{2} x d x = \frac{3}{2}

A = - \int_{- 1}^{0} x d x

+ \int_{0}^{2} x d x

= \frac{1}{2} + 2

= \frac{5}{2}

The sign changes at zero. The negative interval contributes minus one half to the integral but positive one half to total area. The positive interval contributes two to both quantities.

Question 5

Find the real constant that makes this definite integral equal to ten.

\int_{1}^{3} (k x + 1) d x = 10

Show solution 5

{[\frac{k x^{2}}{2} + x]}_{1}^{3}

= 4 k + 2 = 10

\Rightarrow k = 2

Treat k as a constant while antidifferentiating with respect to x. Substitution gives nine halves of k plus three, minus one half of k plus one, leaving four k plus two.

Common questions

Do I include a constant of integration in a definite-integral answer?

An indefinite integral represents a family of antiderivatives, so it needs an arbitrary constant. For a definite integral, the same constant appears in both bound substitutions and cancels. Give the resulting number, or the requested exact expression, rather than appending a constant.

When do I need to split an integral to find total area?

With the lower bound smaller than the upper bound, first locate all zeros inside the interval and check the function’s sign between them. Split wherever the sign changes, then add each region as a positive area. A zero where the curve only touches the axis does not necessarily change the sign. A definite integral adds signed contributions; its absolute value is not generally the total area if cancellation occurs.

Definite integrals

The idea

Worked examples

01 Start with the method

A mistake to avoid

02A nonzero lower bound: subtract a whole expression

03A crossing: signed integral and total area are different

Practice questions with answers

Question 1

Question 2

Question 3

Question 4

Question 5

Common questions

Put it into practice.

Keep exploring.