Class 12 Maths · Original EZT practice
Application of Integrals (Class 12 Maths, NCERT, Chapter 8)
Application of Integrals is Chapter 8 of the NCERT Class 12 Mathematics Part 2 textbook. The area formulas of school geometry stop at polygons and circles - 'they are inadequate for calculating the areas enclosed by curves', and that is what this chapter fixes. Building on Chapter 7's definite integral as a limit of sums, it pictures 'area under the curve as composed of large number of very thin vertical strips', then turns that into method: area under simple curves, and area between lines and arcs of circles, parabolas and ellipses in standard form. Short chapter, one idea, done well. (Note: this is original practice, not a textbook answer key.)
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 26 October 2026:
1. What the chapter covers
8.1 Introduction: why elementary geometry formulas fall short for curved boundaries, and the plan of the chapter. 8.2 Area under Simple Curves: the strip method - dA = y dx; area bounded by y = f(x), the x-axis and ordinates x = a, x = b as the definite integral of f from a to b; areas of circles, parabolas and ellipses in standard forms; area between a line and a curve. EXERCISE 8.1 and the Miscellaneous Exercise on Chapter 8 close the chapter (rationalised edition: the old area-between-two-curves section is not in this printing).
2. Key ideas at a glance
Slice the region into thin vertical strips of height y and width dx; the area is the integral of y dx over the interval. For x = g(y) curves, slice horizontally and integrate with respect to y. Symmetry halves the work: compute one quadrant of a circle or ellipse and multiply. The integral gives signed area; for regions below the x-axis, take the absolute value. Always identify the bounding curves and intersection points before setting limits.
3. Traps this chapter sets
Reading a signed integral as an area: if the curve crosses the axis inside the interval, the integral cancels regions and undercounts - split at the crossing and add absolute values. Starting to integrate before finding intersections: the limits come from where the bounding curves meet, not from the drawing's edge. Forgetting which curve is on top: area between two curves integrates upper minus lower, and that order can swap at an intersection. Skipping symmetry: a full circle or ellipse done directly doubles the risk of a sign slip that a quadrant computation avoids.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Use the complement
Choose the method: Rectangle minus curve.
EZT practice question
The integral of e^x from 0 to 1 gives the area under that curve as e - 1. Without integrating again, find the area of the region bounded by the curve y = e^x, the y-axis and the line y = e.
Show the worked solution
- The rectangle with corners (0, 0), (1, 0), (1,e), (0,e) has area e.
- The curve y = e^x splits it into the area under the curve (e - 1) and the required region above it.
- Required area = e - (e - 1) = 1 square unit.
Answer: 1 square unit - the rectangle's area e minus the given area under the curve (e - 1).
Check: The rectangle corners, the split into two regions, the subtraction.
Watch for this: Do not integrate e^x again or worse, integrate (e - e^x) from scratch - the point of the problem is that the given integral already does the work; complement-region reasoning replaces a second integration.
Now try it yourself
When is complement reasoning available?
Check your answer
Whenever the region and a known integral tile a simple shape - here the curve and the rectangle. If the region does not complete a rectangle (or another known area) with a computed integral, go back to strips.
Spotting the tiling is the skill; the arithmetic is one subtraction.
Slice a circle arc
Choose the method: Chord limits, standard integral.
EZT practice question
Find the area of the region in the first quadrant enclosed by the circle x^2 + y^2 = 9 and the line x = 1.
Show the worked solution
- In the first quadrant y = sqrt(9 - x^2), and the region runs from the line x = 1 to the circle's edge x = 3.
- Area = integral from 1 to 3 of sqrt(9 - x^2) dx.
- Evaluate with the standard formula for sqrt(a^2 - x^2): antiderivative (x/2)sqrt(9 - x^2) + (9/2) arcsin(x/3), giving (9pi/4 - sqrt2 - (9/2)arcsin(1/3)) square units.
Answer: Integral from 1 to 3 of sqrt(9 - x^2) dx, evaluated by the standard sqrt(a^2 - x^2) formula - the chapter's circle-arc pattern, with limits set by the chord x = 1 and the circle's edge x = 3.
Check: First-quadrant branch y = sqrt(9 - x^2), limits 1 and 3, the standard formula.
Watch for this: Do not run the limits from 0 to 3 - the region starts at the line x = 1, not at the y-axis; sketch the chord before choosing limits.
Now try it yourself
Why is x = 3 the upper limit?
Check your answer
The first-quadrant arc meets the x-axis where y = 0, i.e. x = 3 - the region bounded by the circle and the line x = 1 ends at the circle's own edge.
Limits come from the bounding curves, never from habit.
Split at the crossing
Choose the method: Signed first, then absolute.
EZT practice question
The curve y = x^2 - 4 crosses the x-axis inside the interval [0, 3]. Find (a) the definite integral of the curve from 0 to 3, and (b) the true area between the curve and the x-axis over that interval.
Show the worked solution
- Crossing: x^2 - 4 = 0 at x = 2 inside [0, 3].
- (a) Signed integral = [x^3/3 - 4x] from 0 to 3 = (9 - 12) - 0 = -3.
- (b) True area = |integral from 0 to 2| + integral from 2 to 3 = |8/3 - 8| + (9 - 12 - 8/3 + 8) = 16/3 + 7/3 = 23/3 square units.
Answer: (a) -3; (b) 23/3 square units - the signed value and the true area differ whenever the curve dips below the axis inside the interval.
Check: The crossing at x = 2, the split into [0, 2] and [2, 3], the absolute value on the negative part.
Watch for this: Do not report -3 as the area, and do not report the integral of |x^2 - 4| as if it equalled the plain integral - the cancellation between the below-axis and above-axis parts is exactly what the two questions expose.
Now try it yourself
Why is the signed integral negative here even though most of the interval sits above the axis?
Check your answer
The below-axis part on [0, 2] has magnitude 16/3, larger than the 7/3 gained on [2, 3] - the parabola spends the first two units as deep as 4 below the axis.
Signed area is a net balance, not a size.
Upper minus lower
Choose the method: Meet points first.
EZT practice question
Find the area of the region enclosed between the parabola y = x^2 and the line y = x.
Show the worked solution
- Intersections: x^2 = x at x = 0 and x = 1.
- Between them the line lies above the parabola (test x = 1/2: 1/2 > 1/4).
- Area = integral from 0 to 1 of (x - x^2) dx = 1/2 - 1/3 = 1/6 square units.
Answer: 1/6 square units - the integral from 0 to 1 of (line minus parabola), upper curve minus lower curve over the intersection interval.
Check: The two intersections, which curve is on top, upper minus lower.
Watch for this: Do not integrate x^2 - x and take the absolute value blindly - deciding which curve is upper first keeps the integrand positive and the reasoning visible.
Now try it yourself
What changes if the curves cross more than twice?
Check your answer
Each crossing can swap which curve is upper, so the area becomes a sum of integrals, one per interval between consecutive intersections, each with its own upper-minus-lower order.
Intersections partition the problem; the order check is per interval.
Set up, stop
Choose the method: Limits from geometry.
EZT practice question
A region is bounded by the parabola y = x^2 and the line y = 2. Set up (do not evaluate) the integral for its area, explaining your choice of limits.
Show the worked solution
- Intersections: x^2 = 2 at x = -sqrt2 and x = sqrt2.
- Between them the line y = 2 lies above the parabola.
- Area = integral from -sqrt2 to sqrt2 of (2 - x^2) dx - upper curve minus lower curve.
Answer: Area = integral from -sqrt2 to sqrt2 of (2 - x^2) dx; the limits are the x-coordinates where the line meets the parabola, and the integrand is the line minus the parabola because the line is on top between them.
Check: Intersections at plus and minus sqrt2, line above, upper minus lower.
Watch for this: Do not default the limits to 0 and sqrt2 out of symmetry habit without saying why - symmetry would let you write twice the integral from 0 to sqrt2, but the unsaid shortcut is where setup errors hide.
Now try it yourself
Why can a setup-only question be the harder task?
Check your answer
Because evaluation is mechanical once the integral is right; the mathematics of the problem - which curves, which intersections, which order - lives entirely in the setup.
The integral is the answer's body; the setup is its spine.
Slice horizontally
Choose the method: x as a function of y.
EZT practice question
Find the area of the region bounded by the curve x = y^2 and the line x = 4, slicing horizontally.
Show the worked solution
- With x as a function of y, slice with horizontal strips of width dy.
- The curve meets x = 4 at y = -2 and y = 2.
- Area = integral from -2 to 2 of (4 - y^2) dy = [4y - y^3/3] from -2 to 2 = 16/3 + 16/3 = 32/3 square units.
Answer: 32/3 square units - the integral from -2 to 2 of (right curve 4 minus left curve y^2) dy, the x = g(y) slicing pattern of section 8.2.
Check: Horizontal strips, limits -2 and 2, right minus left.
Watch for this: Do not force this into y = sqrt(x) vertical strips - the parabola x = y^2 is not a function of x (two branches), while it is a perfectly good function of y; choose the variable that makes the boundary single-valued.
Now try it yourself
When is horizontal slicing the better choice?
Check your answer
When the boundary is given or more natural as x = g(y), or when vertical strips would need splitting into branches - horizontal strips then give one integral instead of two.
The strip direction is a choice; make it before the algebra starts.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Area under y = f(x) from a to b?
Show answer and reason
The definite integral of f(x) dx from a to b - thin vertical strips of height y, width dx.
dA = y dx is the whole chapter in one line.
2. Curve below the axis?
Show answer and reason
Split at the crossing, take absolute values, then add - the signed integral undercounts.
Area is never negative; integrals can be.
3. Area between two curves?
Show answer and reason
Integrate upper minus lower between consecutive intersection points.
Intersections set the limits and can swap the order.
4. Circle or ellipse area fast?
Show answer and reason
Compute one quadrant and multiply by 4 - symmetry halves the work.
Fewer evaluations, fewer sign slips.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- In two sentences, explain why the definite integral of a function over an interval can be zero while the area between its curve and the x-axis is not.
- Set up (do not evaluate) the area inside the ellipse x^2/25 + y^2/9 = 1 as four times a first-quadrant integral, and state the limits.
- Explain in your own words why the rationalised chapter keeps area between a line and a curve but not the old two-curve section's general formula.
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Questions about this practice page
- Which book is Class 12 Maths Chapter 8 Application of Integrals from?
- It is Chapter 8 of the NCERT Class 12 Mathematics Part 2 textbook (official NCERT chapter PDF lemh202.pdf). No printing-year footer was observed in the fetched chapter text; the front matter lists Edition January 2023 with reprints through 'Reprint 2026-27' - check your own copy, as printings can differ.
- What is the main idea of Application of Integrals?
- Using the definite integral to find areas of regions bounded by curves. The region is imagined as 'large number of very thin vertical strips' of height y and width dx, and the total area is the integral of y dx between the bounding ordinates.
- What is the formula for area under a curve?
- For y = f(x) above the x-axis between x = a and x = b, area A = integral from a to b of f(x) dx. If the curve is given as x = g(y), integrate g(y) dy between the bounding y-values instead.
- Is the area between two curves in this chapter?
- In this rationalised printing, no - the chapter works with area under simple curves and areas between lines and arcs of circles, parabolas and ellipses in standard form. Check your own copy, as contents differ between printings.
- Why does the definite integral sometimes give a negative or zero area?
- The integral is signed: it adds y dx with the sign of y. Where the curve dips below the x-axis the contribution is negative, so compute the area piecewise and take absolute values, or integrate |f(x)|.