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Definite integrals

Evaluate definite integrals with upper minus lower, exact fractions and surds. Solve unknown-limit and parameter questions, and check domains before applying the rule.

Before you startIntegrating powers, substitution and solving equations

01 / An integral between two limits

The order of the limits matters.

An indefinite integral gives a family of antiderivatives. A definite integral accumulates a signed quantity between two inputs. With fixed numeric limits and no remaining parameters, it gives a number.

∫_a^b f(x) dx = F(b) − F(a), where F′ = f

For the continuous functions used here, choose an antiderivative valid throughout the interval. The upper written limit is b; it need not be the larger number.

Change both limits in the model. Positive and negative shaded contributions can cancel, and reversing the limits reverses the sign. This is not automatically a geometric area.

Choose the two limitsMove at your pace

First limit aFinal limit b

For f(x) = 2x − 2 with written limits a = 0 and b = 2, F(b) = 0 and F(a) = 0. The definite integral is 0. Above contributes +; below contributes −. Shading marks the geometric span between the limits; gold is above the axis and blue below it.

Watch the endpoint values produce a definite integral

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Upper minus lower

Substitute into the antiderivative, not the integrand.

Evaluate ∫ from −1 to 2 of 3x² + 2Worked example

∫₋₁²(3x² + 2) dx = [x³ + 2x]₋₁²

Integrate and retain the limits.

= [2³ + 2(2)] − [(−1)³ + 2(−1)]

Put each whole endpoint value in brackets.

= 12 − (−3) = 15

Subtract the lower value, including its minus sign.

Writing f(2) − f(−1) would compare heights, not accumulate the function. The capital F in the evaluation rule is the antiderivative.

03 / Why C cancels

The same constant occurs at both ends.

[F(b) + C] − [F(a) + C] = F(b) − F(a)

You may omit C when evaluating a definite integral because it cancels. This does not justify omitting it from an indefinite integral or a curve-recovery problem.

If a parameter remains in the integrand or limits, the definite integral may be an expression in that parameter. “Definite” does not guarantee a single known numerical value.

04 / Reverse, split and combine

Use consistent endpoints.

∫_a^af(x) dx = 0
∫_b^af(x) dx = −∫_a^bf(x) dx
∫_a^cf(x) dx + ∫_c^bf(x) dx = ∫_a^bf(x) dx

These identities follow by writing each integral as a difference of endpoint values. A constant multiplier can be taken outside, and integrals of sums can be added term by term when the functions are integrable on the interval.

A negative answer may be correct: it can come from a negative integrand, reversed limits or a net balance of positive and negative parts.

What does a vertical shift add to the integral?

∫_a^b[f(x) + c] dx
= ∫_a^bf(x) dx + c(b − a)

If ∫ from 1 to 6 of f(x) is 17, then the integral of f(x) + 4 over the same interval is 17 + 4(6 − 1) = 37. The extra strip has width b − a, not 1. This is an identity for signed integrals; if you are asked for total geometric area, check whether the shift changes any signs or boundaries.

05 / Roots and exact arithmetic

Delay rounding until the final requested answer.

For x > 0:

∫₁⁴(3√x − 2/x²) dx
= [2x^3/2 + 2/x]₁⁴
= (16 + 1/2) − (2 + 2) = 25/2

An exact surd answer

∫₁³(4√x − 1/√x) dx
= [(8/3)x^3/2 − 2√x]₁³
= 6√3 − 2/3

Use 3^3/2 = 3√3. Do not replace √3 by a rounded decimal if an exact answer is requested.

06 / A parameter inside the integral

Integrate with respect to x while holding the parameter fixed.

Suppose ∫ from 0 to 2 of (3Px + 4) equals 2P².

∫₀²(3Px + 4) dx = 6P + 8
6P + 8 = 2P²
P² − 3P − 4 = (P − 4)(P + 1) = 0

Thus P = 4 or P = −1. Both satisfy the stated equation; do not discard a negative parameter without a relevant restriction.

A root integral with a parameter

∫₁⁴(2/√x − A) dx = 4 − 3A
If this equals A²:
A² + 3A − 4 = (A + 4)(A − 1) = 0

The two values are A = −4 and A = 1.

07 / An unknown limit

Evaluate symbolically, then solve and apply the condition.

If k > 1 and ∫ from 1 to k of 2x equals 15:

∫₁^k2x dx = k² − 1
k² − 1 = 15 ⇒ k = ±4
k > 1 selects k = 4

Without the condition k > 1, both −4 and 4 would satisfy this definite-integral equation. The written upper limit is not automatically greater than the lower limit.

A parameter in both limits and integrand

Let k > 0 and suppose ∫ from k to 3k of (x + 2/k) equals 20. Treat k as a constant while integrating in x:

[x²/2 + 2x/k]_k^3k
= (9k²/2 + 6) − (k²/2 + 2)
= 4k² + 4 = 20
k = 2

The positive-domain condition rejects −2.

08 / Check the interval first

A formula cannot integrate through a singularity by accident.

The integrand 1/x² is undefined at zero. Applying [−1/x] from −1 to 1 would give −2, but that calculation crosses a point where the antiderivative is not valid on the whole interval.

There is no finite ordinary integral across that singularity; more advanced improper-integral methods confirm divergence. For this course, flag the domain break instead of reporting the invalid endpoint subtraction.

A student evaluates ∫ from −1 to 1 of 1/x² as −2. What went wrong?

Hint

Inspect the integrand between the endpoints.

Worked solution

It is undefined at zero, so the antiderivative formula cannot be applied across the entire interval. The proposed finite result is invalid.

10 / Recap

Integrate, substitute, subtract, then interpret.

Use an antiderivative valid throughout the relevant interval.
Compute F(b) − F(a) with brackets.
C cancels in a definite integral.
Preserve exact fractions and surds.
Unknown parameters or limits lead to equations and domain checks.
Signed accumulation and geometric area are different quantities.

Next: area under a curve →

Section 1 of 10 · An integral between two limits