WASSCE SHS
Integration reverses differentiation. A definite integral between two limits is a number. An indefinite integral needs a constant.
The integral of axⁿ, for n not equal to −1, is axⁿ⁺¹ divided by the new power, plus a constant if no limits are given. Integrate term by term. The constant matters when the question asks for a curve through a given point: you use the point to find the constant. When limits are given, you do not add a constant. You substitute the upper limit, substitute the lower limit, and subtract.
A definite integral of a rate gives a total change. A definite integral under a curve, where the curve is above the axis, gives an area. If the curve crosses below the axis, areas below come out negative and you handle them as the question asks. The example here stays positive, so the number is the area. Show the antiderivative in brackets with the limits beside it, then the two substitutions. Skipping to a number loses the method marks.
Divide by the new power. Forgetting the division is the common error: the integral of 3x² is x³, not 3x³. For definite integrals, upper minus lower. Swapping the limits changes the sign. The constant of an indefinite integral is part of the answer. Omitting + c loses a mark when no limits were given.
These are practice questions in the style of the paper, with a model answer and a mark scheme. They are not a promise of what your paper will ask.
4 marks
Evaluate the integral of (3x² + 1) with respect to x, from x = 0 to x = 2. (4 marks)
The antiderivative is x³ + x. At 2: 8 + 2 = 10. At 0: 0. Value = 10 − 0 = 10.
2 marks
Find the indefinite integral of 6x. (2 marks)
3x² + c.
3 marks
The gradient of a curve is 6x, and the curve passes through (1, 5). Find the equation of the curve. (3 marks)
y = 3x² + c. At (1, 5): 5 = 3 + c, so c = 2. y = 3x² + 2.
More questions on this topic are inside GLearn, marked as you answer, with the reason written out. The first ones need no account.
Practise Elective Mathematics freeWhen the integral has no limits. The integral of 6x is 3x² + c. When limits are given, you substitute the upper limit and the lower limit and subtract. No constant is written.
Integrate the gradient to get y with an unknown constant, then substitute the point. If the gradient is 6x and the curve passes through (1, 5), then y = 3x² + 2. All WASSCE Elective Mathematics past questions