NCERT Solutions for Class 12 Mathematics Chapter 8: Applications of Integrals – Free PDF Download

Chapter 8 of Class 12 Maths is Applications of Integrals. It uses definite integrals to compute the area under curves, between curves, and bounded by circles/parabolas/ellipses. These Class 12 Mathematics Chapter 8 solutions are also useful as quick revision notes before exams.

Area Under a Curve

The area bounded by y=f(x), the x-axis, and the lines x=a, x=b is Area=∫ab|f(x)|dx (the modulus ensures a positive area even if the curve dips below the axis).

Area Between Two Curves

If f(x)≥g(x) on [a,b], the area between the curves is ∫ab[f(x)−g(x)]dx. Intersection points of the curves are found by solving f(x)=g(x) to determine the limits.

Area of Standard Curves

Circle x²+y²=a²: full area=πa², computed via 4∫0a√(a²−x²)dx for the first quadrant, multiplied by 4. Parabola y²=4ax and ellipse x²/a²+y²/b²=1 areas are computed similarly using integration with symmetry.

Sketching Regions

Before integrating, sketch the curves to correctly identify which function is on top/bottom (or left/right, for integration with respect to y) and determine the correct limits of integration.

Integration with Respect to y

When a region is more naturally described in terms of y (e.g. x=g(y)), the area formula becomes ∫cdx dy=∫cdg(y)dy.

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Frequently Asked Questions

Why is the modulus used in the area formula?
To ensure area is always positive, since integrals below the x-axis are negative but represent real (positive) area.

How do you find the area between two curves?
Find intersection points as limits, then integrate (upper curve − lower curve).

What is the area enclosed by a full circle of radius a?
πa².

Written by Satish

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