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Basic Concept of Definite Integral Part 5

Posted on - 04-01-2017

Qa Math

Bank PO

Inequalities

Sometimes you are asked to prove inequalities involving definite integrals or to estimate the upper and lower bounds of a definite integral, where the exact value of the definite integral is difficult to find. Under these circumstances, we use the following results:.

(i)

Equality sign holds where f (x) is entirely of the same sign on [a, b]

Example.1

Estimate the absolute value of the integral .

Solution

Since |sinx| £ 1 for x ≥ 10, the inequality < 10-8 is fulfilled.

Therefore < (19 – 10) 10-8 < 10-7 (the true value of the integral » -10-8).

(ii)

Example.1

Prove that can’t exceed .

Solution

£

.

(iii) If f (x) ≥ g(x) on [a, b] , then . In particular, if f (x) ≥ 0, then .

Example.1

If f (x) is a continuous function such that f (x) ≥ 0 " x ∈ [2, 10] and

Solution

f(x) is above the x-axis or on the x-axis for all x ∈ [2, 10]. If f (x) is greater than zero for any sub interval of [4, 8], then f (x) d x must be greater than zero. Butf (x) dx = 0


&⇒
f (x) = 0 " x ∈ [4, 8]


&⇒
f (6) = 0

(iv). For a given function f (x) continuous on [a, b] if you are able to find two continuous function f1(x) and f 2(x) on [a, b] such that f1(x) £ f(x) £ f2(x) "x ∈ [a, b], then .

.

Example.1

Prove that

Solution

Since 4 – x2 ≥ 4 – x2 – x3 ≥ 4 – 2x2 > 0 " x ∈ [0, 1]

.

(v). If m and M are the global minimum and global maximum of f (x) respectively in [a, b] then m (b – a)

Example.1

Prove that 4 .

Solution

Since the function f (x) = increases monotonically on the interval [1, 3], m = 2, M = , b – a = 2.

Hence, 2.2
&⇒
4 .

 
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