Class 12 Maths - UP
Application of Integrals
The chapter Application of Integrals in Class 12 Mathematics under the UPMSP curriculum teaches students how to use definite integrals to calculate areas bounded by curves, lines, parabolas, ellipses, and circles. Building directly on the concepts of indefinite and definite integration, this chapter is crucial for board examinations as it tests both visualization skills and algebraic manipulation. Scoring well here requires drawing accurate rough sketches of graphs, finding intersection points, and setting up appropriate vertical or horizontal strip limits. It regularly features high-weightage long-answer questions in the UP Board mathematics paper.
Start Learning FreeKey Concepts
Area under a Curve
The area enclosed between a curve y = f(x), the x-axis, and the ordinates x = a and x = b is given by the definite integral from a to b of f(x) dx.
Area between Two Curves
The area between two intersecting curves y = f(x) and y = g(x) is found by integrating the absolute difference of the upper and lower functions between their intersection points.
Symmetry in Curves
Identifying if a curve is symmetric about the x-axis, y-axis, or origin helps simplify area calculations by integrating over one quadrant and multiplying by a constant.
Integration with respect to y
When curves are more easily expressed as functions of y (x = f(y)), integration along the y-axis between y = c and y = d is used to calculate the area.
Important Formulas
Board Exam Info
In the Uttar Pradesh (UPMSP) Class 12 Mathematics board exam, this chapter typically carries around 6 to 8 marks. Questions usually include one long-answer question (5 or 8 marks) requiring the student to sketch a graph and find the area of a region bounded by a circle, parabola, or line.
Frequently Asked Questions
Is it compulsory to draw a rough sketch in the exam?
Yes, drawing a neat rough sketch showing the region of integration is compulsory and carries steps marks in UP Board exams.
How do we find the limits of integration?
Limits are found by solving the given equations simultaneously to find the x-coordinates or y-coordinates of the points of intersection.
What if the area calculated comes out to be negative?
Definite integrals can yield negative values if the region lies below the x-axis. Since area is always positive, we take the absolute value or use modulus.
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