Chapter 8 Applications of Integrals

Notes

Applications of Integrals ka main use area find karne ke liye hota hai.

Important Concepts:

Area under a curve

If curve y = f(x) is above x-axis from x = a to x = b, then

Area = ∫ₐᵇ f(x) dx

Area between two curves

If y = f(x) is upper curve and y = g(x) is lower curve, then

Area = ∫ₐᵇ [f(x) – g(x)] dx

Area with respect to y-axis

If curves are given as x = f(y), then

Area = ∫ [Right curve – Left curve] dy

Steps to Solve Area Questions:

Find intersection points.

Decide limits.

Identify upper and lower curve.

Apply formula.

Integrate carefully.

Area should always be positive.

Important Formulas:

Area under y = f(x):

Area = ∫ₐᵇ y dx

Area between y = f(x) and y = g(x):

Area = ∫ₐᵇ (Upper curve – Lower curve) dx

Area between x = f(y) and x = g(y):

Area = ∫ (Right curve – Left curve) dy

Important Standard Integrals:

∫ xⁿ dx = xⁿ⁺¹/(n + 1)

∫ sin x dx = –cos x

∫ cos x dx = sin x

∫ eˣ dx = eˣ

∫ 1/x dx = log |x|

Useful Tricks:

Area can never be negative.

If answer comes negative, take positive value.

For symmetric graph, calculate half area and multiply by 2.

For parabola and line questions, first find intersection points.

Always write upper curve minus lower curve.

Common Mistakes:

Wrong limits.

Wrong upper and lower curve.

Forgetting to find intersection points.

Taking negative area.

Calculation mistake in definite integral.

Confusion between dx and dy.

Solved Example:

Find area between y = x and y = x².

Solution:

x = x²

x(x – 1) = 0

x = 0, 1

Area = ∫₀¹ (x – x²) dx

= [x²/2 – x³/3]₀¹

= 1/2 – 1/3

= 1/6

Answer:

1/6 square units