Q. 71 Prove聽鈭玿kdx=1k+1xk+1+C聽if聽k... [FREE SOLUTION]

Chapter 4: Q. 71 (page 363)

Prove $鈭� x^{k} d x = \frac{1}{k + 1} x^{k + 1} + C i f k 鈭� 1 a n d 鈭� \frac{1}{x} d (x) = \ln (x) + C$

Short Answer

Expert verified

Hence Proved

Step by step solution

Step 1. Given

The proving equation is $鈭� x^{k} d x = \frac{1}{k + 1} x^{k + 1} + C i f k 鈭� 1 a n d 鈭� \frac{1}{x} d (x) = \ln (x) + C$ .

Step 2. Finding derivative

$Finding derivative of \frac{1}{k + 1} x^{k + 1} d (\frac{1}{k + 1} x^{k + 1}) = \frac{1}{k + 1} (k + 1) x^{k + 1 - 1} = x^{k} Since \frac{1}{k + 1} x^{k + 1} is antiderivative of x^{k} Therefore 鈭� x^{k} d (x) = \frac{1}{k + 1} x^{k + 1}$

Step 3. Proving

$鈭� \frac{1}{x} d (x) = \ln (x) + C N o w, \frac{d}{d x} (\ln (x)) = \frac{1}{x} S i n c e, \ln (x) i s a n a n t i d e r v a t i v e o f \frac{1}{x} T h e r e f o r e 鈭� \frac{1}{x} = \ln (x) + C$

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91影视

Prove $鈭� x^{k} d x = \frac{1}{k + 1} x^{k + 1} + C i f k 鈭� 1 a n d 鈭� \frac{1}{x} d (x) = \ln (x) + C$

Short Answer

Step by step solution

Step 1. Given

Step 2. Finding derivative

Step 3. Proving

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91影视

Prove鈭�xkdx=1k+1xk+1+Cifk鈭�1and鈭�1xd(x)=ln(x)+C

Short Answer

Step by step solution

Step 1. Given

Step 2. Finding derivative

Step 3. Proving

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Applied Mathematics

Calculus

Discrete Mathematics

Geometry

Decision Maths

Study anywhere. Anytime. Across all devices.

Prove $鈭� x^{k} d x = \frac{1}{k + 1} x^{k + 1} + C i f k 鈭� 1 a n d 鈭� \frac{1}{x} d (x) = \ln (x) + C$