Geometrical Interpretation of Definite Integral

Important PropertiesHow to Measure Definite IntegralSolved Problems on Definite Integral

Consider a curve which is above the x-axis. It is a continuous function on an interval [a, b] where all the values are positive. The area between the curve and the x-axis defines the definite integral

of any continuous function. In the above formulae, a and b are the limits, d/dx (F(x)) = f (x).

The definite integral is different from the indefinite integral, as follows:

• Indefinite integral lays the base for definite integral.
• Indefinite integral defines the calculation of indefinite area, whereas definite integral is finding the area with specified limits.

Important Properties

The following are the properties of definite integrals:

1. ∫_a^b f (x) dx = ∫_a^b f (t) dt

2. ∫_a^b f (x) dx = – ∫_b^a f (x) dx … [Also, ∫_a^a f(x) dx = 0]

3. ∫_a^b f (x) dx = ∫_a^c f (x) dx + ∫_c^b f (x) dx

4. ∫_a^b f (x) dx = ∫_a^b f (a + b – x) dx

5. ∫₀^a f (x) dx = ∫₀^a f (a – x) dx …

6. ∫₀^2a f (x) dx = ∫₀^a f (x) dx + ∫₀^a f (2a – x) dx

7. Two parts

• ∫₀^2a f (x) dx = 2 ∫₀^a f (x) dx … if f (2a – x) = f (x).
• ∫₀^2a f (x) dx = 0 … if f (2a – x) = – f(x)

8. Two parts

1. ∫_-a^a f(x) dx = 2 ∫₀^a f(x) dx … if f(- x) = f(x) or it is an even function
2. ∫_-a^a f(x) dx = 0 … if f(- x) = – f(x) or it is an odd function

9. Walli’s formula

10. Leibnitz’s rule

If g is continuous on [a, b], and f1 (x) and f2 (x) are the two differentiable functions whose values lie in [a, b], then d / dx ∫ f2(x) f1(x) g(t) dt = g (f2(x)) f2′(x) – g (f1(x)) f1′(x)

How to Measure Definite Integral

The area under the graph is the definite integral. By definition, definite integral is the sum of the product of the lengths of intervals and the height of the function that is being integrated with that interval, which includes the formula of the area of the rectangle. The figure given below illustrates it.

Solved Problems on Definite Integral

Example 1: ∫₀^π/2 ([√cotx] / [√cotx + √tanx]) dx

Solution:

I = ∫₀^π/2 ([√cotx] / [√cotx + √tanx]) dx ….. (i)

= ∫₀^π/2 (√[cot (π / 2 − x)] / √cot(π / 2 − x) + √tan (π / 2 − x) ) dx

= ∫₀^π/2 ([√tanx] / [√tanx + √cotx]) dx ….. (ii)

Now adding (i) and (ii), we get

2I = ∫₀^π/2 ( [√cotx + √tanx / [√tanx + √cotx]) dx

=[x]₀^π/2 ⇒ I = π / 4

Example 2: ∫₀^π/2 dθ / [1 + tanθ]

Solution:

I = ∫₀^π/2 dθ / [1 + tanθ]

= ∫₀^π/2 dθ / [1 + tan (π / 2 − θ)]

=∫₀^π/2 dθ / [1 + cotθ]

On adding,

2I = ∫₀^π/2(1 / [1 + tanθ]) + (1 / [1 + cotθ]) dθ

=∫₀^π/2 dθ = [θ]₀^π/2

= π / 2 ⇒I = π / 4

Example 3: ∫₀^π/2 ([cosx−sinx] / [1 + sinx cosx]) dx

Solution:

I = ∫₀^π/2 ([cosx−sinx] / [1 + sinx cosx]) dx ……(i)

Now I = ∫₀^π/2 cos (π / 2 − x) − sin (π / 2 − x) 1 + sin (π / 2− x) cos (π / 2 − x) dx = ∫₀^π/2 ([sinx − cosx] / [1 + sinx cosx]) dx ……(ii)

On adding, 2I = 0 ⇒ I = 0

Example 4: ∫_-1¹ (log [2−x] / [2+x]) dx

Solution:

Let f (x) = (log [2−x] / [2+x])

⇒f (−x) = (log [2−x] / [2+x]) ^-1

= − (log [2−x] / [2+x]) = −f(x)

∴ ∫_-1¹ (log [2−x] / [2+x]) dx = 0

Example 5: What is the smallest interval [a,b], such that ∫₀¹ dx / √[1 + x⁴] ∈ [a, b]?

Solution:

Let I = ∫₀¹ dx / √[1 + x⁴]

Here, 0 ≤ x ≤1 ⇒ 1 ≤ (1 + x⁴) ≤ 2

1 ≤ √[1 + x⁴] ≤ √2 ⇒ 1 / √2 ≤ 1 / √[1 + x⁴] ≤1

1 / √2 ≤ ∫₀¹ dx / √[1 + x⁴] ≤ 1

Hence, [1 / √2, 1] is the smallest interval, such that I ∈ [1 / √2, 1].

Note: If m = least value of f (x) and M = greatest value of f (x) in [a, b], then

m (b−a) ≤ ∫_a^b f (x) dx ≤ M (b − a)

Example 6: If n is a positive integer and [x] is the greatest integer not exceeding x, then

∫₀ⁿ {x − [x]} dx equals _____________.

Solution:

x − [x] is a periodic function with period 1.

∴∫₀ⁿ {x − [x]} dx = ∫₀ⁿ (x − [x]) dx

= n [∫₀¹ x dx −∫₀¹ [x] dx]

= n [(x² / 2)₀¹ − 0] = n / 2

Example 7:

Solution: 

Putting [2x = t]

Example 8:

Solution: 

Adding (i) and (ii),

Example 9:

Solution:

Example 10:

Solution: