Q. 79 Prove that for all real numbers ... [FREE SOLUTION]

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Chapter 4: Q. 79 (page 387)

Prove that for all real numbers a and b with a < b, we have $|{鈭�}_{a}^{b} f (x) d x| 猢� {鈭�}_{a}^{b} |f (x)| d x .$

Short Answer

Expert verified

Hence, proved.

Step by step solution

Step 1. Given Information.

a and b are real numbers and a<b.

Step 2. Modulus property.

From the modulus property we know that,

$- |f (x)| 猢� f (x) 猢� |f (x)|, A p p l y i n g i n t e g r a t i o n o v e r t h i s i n e q u a l i t y w e g e t, - {鈭�}_{a}^{b} |f (x)| d x 猢� {鈭�}_{a}^{b} f (x) d x 猢� {鈭�}_{a}^{b} |f (x)| d x . N o w f r o m t h i s w e n e e d o n l y, {鈭�}_{a}^{b} f (x) d x 猢� {鈭�}_{a}^{b} |f (x)| d x .$

Step 3. Proof.

As we all know,

$|a| 猢� a, T h e r e f o r e l e t a = {鈭�}_{a}^{b} f (x) d x . T h i s m e a n s, |{鈭�}_{a}^{b} f (x) d x| 猢� {鈭�}_{a}^{b} f (x) d x . A n d a l s o f r o m s t e p 2 w e h a v e, {鈭�}_{a}^{b} f (x) d x 猢� {鈭�}_{a}^{b} |f (x)| d x . C o m b i n i n g a b o v e t w o r e s u l t s w e g e t, {鈭�}_{a}^{b} |f (x)| d x 猢� |{鈭�}_{a}^{b} f (x) d x| . H e n c e, P r o v e d .$

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91影视

Prove that for all real numbers a and b with a < b, we have $|{鈭�}_{a}^{b} f (x) d x| 猢� {鈭�}_{a}^{b} |f (x)| d x .$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Modulus property.

Step 3. Proof.

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91影视

Prove that for all real numbers a and b with a < b, we have鈭�abf(x)dx猢�鈭�abf(x)dx.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Modulus property.

Step 3. Proof.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Decision Maths

Theoretical and Mathematical Physics

Applied Mathematics

Mechanics Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

Prove that for all real numbers a and b with a < b, we have $|{鈭�}_{a}^{b} f (x) d x| 猢� {鈭�}_{a}^{b} |f (x)| d x .$