The definite integral
Evaluate definite integrals using the fundamental theorem of calculus; interpret the definite integral as the signed area under a curve; work with regions above and below the -axis.
Worked examples
Evaluating a definite integral
Straightforward
Problem
Evaluate .
1
Find the antiderivative.
2
Apply the limits using the notation .
3
Calculate each value.
4
Subtract.
Answer
Signed area and regions below the axis
Moderate
Problem
Evaluate and explain the result geometrically.
1
Find the antiderivative and apply the limits.
2
Interpret the result geometrically.
The definite integral is zero, but the curve is not zero everywhere. On , (area above the axis), and on , (area below the axis). The two areas are equal in magnitude, so they cancel:
Key point: the definite integral gives signed area. To find the total enclosed area, integrate on each sub-interval separately and add the absolute values.
Answer
, since the equal positive and negative signed areas cancel.
Applying the Fundamental Theorem of Calculus
Challenging
Problem
Define . Find and hence find .
1
Apply the Fundamental Theorem of Calculus: if , then .
2
Evaluate at .
Note that the lower limit plays no role in the derivative — only the integrand evaluated at the upper limit matters.
Answer
and .
Practise
Q1·Straightforward
Evaluate .
Explanation
Q2·Straightforward
Evaluate , correct to two decimal places.
Explanation
Rounded to two decimal places: .
Q3·Straightforward
Evaluate .
Explanation
Q4·Straightforward
Evaluate .
Explanation
Q5·Straightforward
Evaluate , correct to two decimal places.
Explanation
Rounded to two decimal places: .
Q6·Moderate
Evaluate , correct to two decimal places.
Explanation
Q7·Moderate
Evaluate . Give your answer as a fraction.
/
Explanation
Q8·Moderate
Evaluate . Give your answer as a fraction.
/
Explanation
Q9·Moderate
The region between and the -axis for lies entirely above the -axis. Find its area. Give your answer as a fraction.
/
Explanation
Since the parabola is non-negative on , the area equals the definite integral:
Q10·Moderate
Evaluate .
Explanation
The negative value reflects that more of the curve lies below than above it on .
Q11·Challenging
Evaluate , correct to two decimal places.
Explanation
Rewrite the integrand by dividing each term by :
Now integrate:
Rounded to two decimal places: .
Now integrate:
Rounded to two decimal places: .
Q12·Challenging
Define . Find .
Explanation
By the Fundamental Theorem of Calculus:
Therefore:
Therefore:
Q13·Challenging
Find the total area enclosed between the curve and the -axis for . Give your answer as a fraction.
*Note: the curve crosses the -axis within this interval, so you will need to handle each sub-region separately.*
*Note: the curve crosses the -axis within this interval, so you will need to handle each sub-region separately.*
/
Explanation
Factorise: , which has zeros at .
On : Check : , so the curve is above the -axis.
On : Check : , so the curve is below the -axis.
Total area
On : Check : , so the curve is above the -axis.
On : Check : , so the curve is below the -axis.
Total area
Open Math
The definite integral
Calculus · MAV-12-05
Name:
Date:
Q1Straightforward
Evaluate .
Q2Straightforward
Evaluate , correct to two decimal places.
Q3Straightforward
Evaluate .
Q4Straightforward
Evaluate .
Q5Straightforward
Evaluate , correct to two decimal places.
Q6Moderate
Evaluate , correct to two decimal places.
Q7Moderate
Evaluate . Give your answer as a fraction.
Q8Moderate
Evaluate . Give your answer as a fraction.
Q9Moderate
The region between and the -axis for lies entirely above the -axis. Find its area. Give your answer as a fraction.
Q10Moderate
Evaluate .
Q11Challenging
Evaluate , correct to two decimal places.
Q12Challenging
Define . Find .
Q13Challenging
Find the total area enclosed between the curve and the -axis for . Give your answer as a fraction.
*Note: the curve crosses the -axis within this interval, so you will need to handle each sub-region separately.*
*Note: the curve crosses the -axis within this interval, so you will need to handle each sub-region separately.*
Worked solutions and answers at openmath.au/year-12/advanced/integral-calculus/the-definite-integral