Integration
Algebra · Integration
Syllabus tag: KCSE | Mathematics | Form 4 | Topic 10 Integration
Lesson objectives
By the end of this topic, you should be able to:
- Carry out the process of differentiation in reverse.
- Determine the constant of integration and find indefinite integrals.
- Evaluate definite integrals.
- Apply integration to find areas under curves and to solve kinematics problems.
Integration
Integration reverses differentiation. Where differentiation finds a gradient function, integration recovers the original from it.
a) The reverse process
Differentiating multiplies by the power and reduces it by one.
Integrating therefore raises the power by one and divides by the new power.
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, provided n is not −1.
Check any integration by differentiating your answer. It must return the original.
b) The constant of integration
Every constant differentiates to zero. So x² + 3, x² − 7 and x² all have the same derivative, 2x.
Reversing 2x therefore cannot recover which constant was there.
We write + c to acknowledge that. This is the constant of integration, and omitting it from an indefinite integral loses marks.
c) Finding the constant
If one point on the curve is known, substitute it to find c.
Say dy/dx = 2x and the curve passes through (1, 5). Integrating gives y = x² + c. Then 5 = 1 + c, so c = 4. The curve is y = x² + 4.
d) Definite integrals
A definite integral has limits and gives a number, not a function.
Integrate, then substitute the upper limit and subtract the value at the lower limit.
The constant cancels in the subtraction, which is why definite integrals need no + c.
e) Area under a curve
The definite integral gives the area between the curve and the x-axis, between the limits.
Where the curve lies below the axis, the integral is negative.
Suppose a question asks for a physical area and the curve crosses the axis. Split the integral at the crossing and add the absolute values. Signed values let positive and negative regions cancel.
f) Area between two curves
Integrate the difference of the two functions, upper minus lower, between the points where they meet.
Find the intersections first, since those are the limits.
g) Kinematics
Integration reverses the differentiation chain.
Integrating acceleration gives velocity. Integrating velocity gives displacement.
Each integration introduces a constant, found from the initial conditions, such as the velocity at t = 0.
Note that the definite integral of velocity gives displacement, not distance. If the body reverses direction, split at that moment and add the absolute values for distance.
h) Where this is used
Areas and volumes of irregular shapes. Total distance from a velocity record. Work done from a force. Accumulating any quantity given its rate.
Words to know
- Integration -- the reverse process of differentiation.
- Constant of integration -- the unknown constant c in an indefinite integral.
- Indefinite integral -- an integral without limits, giving a function plus c.
- Definite integral -- an integral between limits, giving a number.
- Limits -- the two values between which a definite integral is evaluated.
:::checkpoint Check yourself
- Find ∫ 6x² dx.
- Why is a constant of integration needed?
- Evaluate the definite integral of 3x² from 0 to 2.
- What does integrating acceleration give? :::
Bridge to practice
The exercises begin with the rule and the constant of integration, move through definite integrals and areas, and finish with areas between graphs and kinematics. For every area question, sketch the curve first and check whether it crosses the x-axis within the limits.