Integration (Indefinite and Definite)

Learn indefinite and definite integration at A-Level: the power rule for integration, finding areas under curves, and the constant of integration.

Integration is one of the two central operations of calculus, alongside differentiation. At AS Level, integration is introduced as the reverse process of differentiation, and you will learn to evaluate both indefinite and definite integrals of polynomial functions. This topic is tested by all major exam boards — AQA, Edexcel, and OCR — and forms the basis for more advanced integration techniques studied at A2.

The key ideas at this stage are: finding a function whose derivative is a given function (indefinite integration), calculating the exact area bounded by a curve and the xx-axis (definite integration), and understanding the role of the constant of integration. These skills connect directly to real-world applications in physics, engineering, and economics, where quantities such as displacement, area, and accumulated totals are computed via integration.

In this guide, we cover the fundamental rules, work through detailed examples, and highlight the common pitfalls that trip up students in examinations.

Core Concepts

Integration as the Reverse of Differentiation

If ddx[F(x)]=f(x)\frac{d}{dx}[F(x)] = f(x), then we say that F(x)F(x) is an antiderivative (or integral) of f(x)f(x). We write:

f(x)dx=F(x)+c\int f(x) \, dx = F(x) + c

where cc is the constant of integration. This constant arises because the derivative of any constant is zero, so infinitely many functions share the same derivative.

For example, since ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2, we have 3x2dx=x3+c\int 3x^2 \, dx = x^3 + c.

The Power Rule for Integration

The most fundamental integration rule at AS Level is:

xndx=xn+1n+1+c(n1)\int x^n \, dx = \frac{x^{n+1}}{n+1} + c \quad (n \neq -1)

This works by reversing the power rule for differentiation. You increase the power by one and divide by the new power.

For a general term axnax^n:

axndx=axn+1n+1+c\int ax^n \, dx = \frac{ax^{n+1}}{n+1} + c

Integration is linear, meaning you can integrate a sum or difference of terms by integrating each term separately:

[f(x)+g(x)]dx=f(x)dx+g(x)dx\int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx

Definite Integrals

A definite integral has upper and lower limits of integration and produces a numerical value:

abf(x)dx=[F(x)]ab=F(b)F(a)\int_a^b f(x) \, dx = \left[ F(x) \right]_a^b = F(b) - F(a)

The constant of integration cancels when evaluating a definite integral, so we do not include +c+c.

The Fundamental Theorem of Calculus guarantees that this process correctly computes the signed area between the curve y=f(x)y = f(x) and the xx-axis, from x=ax = a to x=bx = b.

Area Under a Curve

The definite integral abf(x)dx\int_a^b f(x) \, dx gives the signed area between the curve and the xx-axis. When f(x)0f(x) \geq 0 on [a,b][a, b], the integral equals the area directly. When f(x)<0f(x) < 0 on part of the interval, the integral gives a negative contribution for that region.

To find the total area (always positive), you must:

  1. Find where f(x)=0f(x) = 0 (the roots) within the interval.
  2. Split the integral at these roots.
  3. Take the absolute value of each part and add them together.

Preparing Expressions for Integration

Before integrating, you may need to rewrite expressions so that every term is in the form axnax^n. Common manipulations include:

  • Expanding brackets: x(x+3)dx=(x2+3x)dx\int x(x + 3) \, dx = \int (x^2 + 3x) \, dx
  • Dividing terms: x3+2xxdx=(x2+2)dx\int \frac{x^3 + 2x}{x} \, dx = \int (x^2 + 2) \, dx
  • Rewriting roots: xdx=x1/2dx\int \sqrt{x} \, dx = \int x^{1/2} \, dx
  • Rewriting reciprocals: 3x2dx=3x2dx\int \frac{3}{x^2} \, dx = \int 3x^{-2} \, dx

Finding the Constant of Integration

When given additional information (such as a point the curve passes through), you can determine the value of cc. If f(x)dx=F(x)+c\int f(x) \, dx = F(x) + c and the curve passes through (a,b)(a, b), then b=F(a)+cb = F(a) + c, giving c=bF(a)c = b - F(a).

Strategy Tips

Tip 1: Always Include +c+c for Indefinite Integrals

Forgetting the constant of integration is one of the most common mark losses. For indefinite integrals, always write +c+c at the end. For definite integrals (with limits), you do not need +c+c.

Tip 2: Rewrite Before Integrating

If the integrand is not already a sum of terms like axnax^n, rewrite it first. Expand products, divide fractions term by term, and convert roots and reciprocals to index notation before applying the power rule.

Tip 3: Use Square Brackets with Limits for Definite Integrals

When evaluating abf(x)dx\int_a^b f(x) \, dx, write [F(x)]ab\left[ F(x) \right]_a^b and then substitute the upper limit first, minus the lower limit. This notation scores method marks and keeps your working clear.

Tip 4: Check by Differentiating

After integrating, differentiate your answer. If you get back to the original integrand, your integration is correct. This is the most reliable self-check available.

Tip 5: Handle Negative Areas Carefully

When asked for an "area", remember that the definite integral gives signed area. If the curve dips below the xx-axis, split the integral and take absolute values. A sketch of the curve helps you identify where f(x)<0f(x) < 0.

Worked Example: Example 1

Problem

Find (3x2+4x5)dx\int (3x^2 + 4x - 5) \, dx.

Solution

Integrate term by term using the power rule:

3x2dx=3x33=x3\int 3x^2 \, dx = \frac{3x^3}{3} = x^3

4xdx=4x22=2x2\int 4x \, dx = \frac{4x^2}{2} = 2x^2

(5)dx=5x\int (-5) \, dx = -5x

Combining:

(3x2+4x5)dx=x3+2x25x+c\int (3x^2 + 4x - 5) \, dx = x^3 + 2x^2 - 5x + c

Worked Example: Example 2

Problem

Evaluate 13(2x+1)dx\int_1^3 (2x + 1) \, dx.

Solution

13(2x+1)dx=[x2+x]13\int_1^3 (2x + 1) \, dx = \left[ x^2 + x \right]_1^3

Substitute the upper limit x=3x = 3: (3)2+3=12(3)^2 + 3 = 12

Substitute the lower limit x=1x = 1: (1)2+1=2(1)^2 + 1 = 2

=122=10= 12 - 2 = 10

Worked Example: Example 3

Problem

Find the area enclosed between the curve y=x24xy = x^2 - 4x and the xx-axis.

Solution

First, find where the curve crosses the xx-axis: x24x=0x^2 - 4x = 0, so x(x4)=0x(x - 4) = 0, giving x=0x = 0 and x=4x = 4.

Between x=0x = 0 and x=4x = 4, the curve is below the xx-axis (you can check: at x=2x = 2, y=48=4<0y = 4 - 8 = -4 < 0).

04(x24x)dx=[x332x2]04\int_0^4 (x^2 - 4x) \, dx = \left[ \frac{x^3}{3} - 2x^2 \right]_0^4

=(64332)(00)=643963=323= \left( \frac{64}{3} - 32 \right) - (0 - 0) = \frac{64}{3} - \frac{96}{3} = -\frac{32}{3}

Since the area is below the axis, the integral is negative. The actual area is:

Area=323=323 square units\text{Area} = \left| -\frac{32}{3} \right| = \frac{32}{3} \text{ square units}

Worked Example: Example 4

Problem

A curve has gradient function dydx=6x22x+1\frac{dy}{dx} = 6x^2 - 2x + 1 and passes through the point (1,8)(1, 8). Find the equation of the curve.

Solution

y=(6x22x+1)dx=2x3x2+x+cy = \int (6x^2 - 2x + 1) \, dx = 2x^3 - x^2 + x + c

Using the point (1,8)(1, 8):

8=2(1)3(1)2+1+c=21+1+c=2+c8 = 2(1)^3 - (1)^2 + 1 + c = 2 - 1 + 1 + c = 2 + c

So c=6c = 6, and the equation of the curve is y=2x3x2+x+6y = 2x^3 - x^2 + x + 6.

Worked Example: Example 5

Problem

Find x3+4x2dx\int \frac{x^3 + 4}{x^2} \, dx.

Solution

Rewrite by dividing each term by x2x^2:

x3+4x2=x+4x2\frac{x^3 + 4}{x^2} = x + 4x^{-2}

Now integrate:

(x+4x2)dx=x22+4x11+c=x224x+c\int (x + 4x^{-2}) \, dx = \frac{x^2}{2} + \frac{4x^{-1}}{-1} + c = \frac{x^2}{2} - \frac{4}{x} + c

Practice Problems

  1. Problem 1

    Find (5x43x2+7)dx\int (5x^4 - 3x^2 + 7) \, dx. [Answer: x5x3+7x+cx^5 - x^3 + 7x + c]

    Problem 2

    Evaluate 02(x33x)dx\int_0^2 (x^3 - 3x) \, dx. [Answer: 2-2]

    Problem 3

    Find the total area enclosed between the curve y=x21y = x^2 - 1 and the xx-axis between x=1x = -1 and x=2x = 2. [Answer: 43+23+23=2\frac{4}{3} + \frac{2}{3} + \frac{2}{3} = 2 — split at roots x=±1x = \pm 1]

    Problem 4

    A curve has dydx=4x3\frac{dy}{dx} = 4x - 3 and passes through (2,5)(2, 5). Find the equation of the curve. [Answer: y=2x23x+3y = 2x^2 - 3x + 3]

    Problem 5

    Find (x+1x)dx\int (\sqrt{x} + \frac{1}{\sqrt{x}}) \, dx. [Answer: 23x3/2+2x1/2+c\frac{2}{3}x^{3/2} + 2x^{1/2} + c]

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Common Mistakes

  • Forgetting +c+c on indefinite integrals. This is marked separately on most exam mark schemes. Even if your integration is perfect, omitting +c+c loses a mark.

  • Incorrect application of the power rule. Students sometimes forget to divide by the new power. For instance, x3dx=x44+c\int x^3 \, dx = \frac{x^4}{4} + c, not x4+cx^4 + c.

  • Not rewriting expressions before integrating. You cannot integrate x3+4x2\frac{x^3+4}{x^2} directly. It must first be split into x+4x2x + 4x^{-2} before the power rule applies.

  • Sign errors in definite integrals. When substituting limits, be careful with negative values. Compute F(b)F(b) and F(a)F(a) separately, then subtract.

  • Confusing area with definite integral. The definite integral gives signed area. If the question asks for "the area", you need the magnitude (absolute value) of each section, particularly when the curve crosses the xx-axis.

  • Integrating x1x^{-1} using the power rule. The formula xndx=xn+1n+1+c\int x^n \, dx = \frac{x^{n+1}}{n+1} + c does not apply when n=1n = -1 (as it would involve division by zero). The integral of 1x\frac{1}{x} is lnx+c\ln|x| + c, which is covered at A2.

Frequently Asked Questions

Why do we add a constant of integration?

Because the derivative of a constant is zero, many different functions can have the same derivative. For example, x2+5x^2 + 5 and x23x^2 - 3 both have derivative 2x2x. The constant cc accounts for all possible antiderivatives.

Do I need $+c$ on definite integrals?

No. When you evaluate a definite integral using F(b)F(a)F(b) - F(a), the constant cancels: (F(b)+c)(F(a)+c)=F(b)F(a)(F(b) + c) - (F(a) + c) = F(b) - F(a). So the +c+c is unnecessary for definite integrals.

Can I check my integration?

Yes — differentiate your answer. If differentiating F(x)+cF(x) + c gives you back the original f(x)f(x), your integration is correct. This is the most reliable verification method.

How do I know when to use definite vs indefinite integration?

If the question gives you limits of integration or asks for an area, a specific value, or a numerical result, use definite integration. If it asks you to "find" or "determine" a general expression (often as part of finding a curve equation), use indefinite integration with +c+c.

What happens if I get a negative answer for a definite integral?

A negative definite integral means the curve is below the xx-axis over that interval. If the question asks for the integral, write the negative value. If it asks for the area, take the absolute value.

Key Takeaways

  • Integration reverses differentiation. If differentiating gives you f(x)f(x), integrating f(x)f(x) recovers the original function (plus a constant).

  • The power rule is your main tool. xndx=xn+1n+1+c\int x^n \, dx = \frac{x^{n+1}}{n+1} + c for n1n \neq -1. Increase the power by one and divide by the new power.

  • Always include the constant of integration. For indefinite integrals, +c+c is mandatory. It represents the family of all antiderivatives.

  • Definite integrals give signed area. Use the Fundamental Theorem: abf(x)dx=F(b)F(a)\int_a^b f(x) \, dx = F(b) - F(a). For total area, split at roots and take absolute values.

  • Rewrite before integrating. Expand brackets, simplify fractions, and convert roots to index form so every term is in the form axnax^n.

  • Check by differentiating. The simplest and most effective way to verify your integration is to differentiate the result and confirm you recover the original function.

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