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Chapter 4: Q. 73 (page 363)

Prove (a) $鈭� e^{k x} d x = \frac{1}{k} e^{k x} + C using differentiation formulas$
(b) $鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C by differentiation formulas$

Short Answer

Expert verified

Hence proved

Step by step solution

01

Step 1. To prove

(a) $鈭� e^{k x} d x = \frac{1}{k} e^{k x} + C$

(b) $鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C$

02

Part(a) Step 2. Proving

$鈭� e^{k x} d x = \frac{1}{k} e^{k x} + C N o w, \frac{d}{d x} (\frac{1}{k}) e^{k x} = (\frac{1}{k}) k e^{k x} = e^{k x} s o, \frac{1}{k} e^{k x} is an antiderivative of e^{kx} Hence proved$

03

Part( b) Step 3. Proving

$鈭� b^{x} d x = \frac{1}{\ln b} b^{x} + C \frac{d}{d x} (\frac{1}{\ln b} b^{x}) = (\frac{1}{\ln b} b^{x}) \ln b = b^{x} \frac{1}{\ln b} b^{x} is an antiderivative of b^{x} . Hence proved$

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Most popular questions from this chapter

Consider the general sigma notation ${鈭�}_{k = m}^{n} a_{k}$ .What do we mean when we say that a_k is a function of k?

Use a sentence to describe what the notation ${鈭�}_{k = 3}^{87} k^{2}$ means. (Hint: Start with 鈥淭he sum of....鈥�)

Verify that $鈭� \ln x d x = x (\ln x - 1) + C$ (Do not try to solve the integral from scratch.

Write out all the integration formulas and rules that we know at this point.

Use integration formulas to solve each integral in Exercises 21鈥�62. You may have to use algebra, educated guess- and- check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. (Hint for Exercise 54: $\tan (x) = \frac{\sin (x)}{\cos (x)}$ ).

$鈭� \frac{e^{3 x} - 2 e^{4 x}}{e^{2 x}} d x$ .

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