Chapter 4: Q. 72 (page 363)

Prove $鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� e^{k x} d x = \frac{1}{k} e^{k x} + C$
and $鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C$

Short Answer

Expert verified

Hence Proved.

Step by step solution

Step 1. To prove

(a) $鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� e^{k x} d x = \frac{1}{k} e^{k x} + C$

(b)role="math" localid="1648664298132" $鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C$

Part (a) Step 2. Proving

$鈭� e^{x} d x = e^{x} + C by using 鈭� e^{k x} d x = \frac{1}{k} e^{k x} + C = 鈭� e^{x} d x = \frac{1}{k} e^{k x} + C = 鈭� e^{x (1)} d x = \frac{1}{k} e^{x (1)} + C = 鈭� e^{x} d x = e^{x} + C Hence proved$

Part( b) Step 3. Proving

$鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C = 鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C = 鈭� e^{x} d x = \frac{1}{\ln e} e^{x} + C = 鈭� e^{x} d x = \frac{1}{1} e^{x} + C = 鈭� e^{x} d x = e^{x} + C Hence proved$

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Prove $鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� e^{k x} d x = \frac{1}{k} e^{k x} + C$
and $鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C$

Short Answer

Step by step solution

Step 1. To prove

Part (a) Step 2. Proving

Part( b) Step 3. Proving

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Prove 鈭�exdx=ex+Cbyusing鈭�ekxdx=1kekx+Cand鈭�exdx=ex+Cbyusing鈭�bxdx=1lnbbx+C

Short Answer

Step by step solution

Step 1. To prove

Part (a) Step 2. Proving

Part( b) Step 3. Proving

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

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Prove $鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� e^{k x} d x = \frac{1}{k} e^{k x} + C$
and $鈭� e^{x} d x = e^{x} + C b y u \sin g 鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C$