If $I = \sum_{k=1}^{98} \int_{k}^{k+1} \frac{dx}{x(x+1)}$, then Options: (A) $I > \log_e 99$ (B) $I \frac{49}{50}$

### Analyzing the Setup The expression provided is: $$I = \sum_{k=1}^{98} \int_{k}^{k+1} \frac{dx}{x(x+1)}$$ The key to solving this is to decompose the rational function $\frac{1}{x(x+1)}$. Whenever you see a product in the denominator, your first instinct should be **partial fraction decomposition**. We can rewrite the integrand as: $$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$$ This transformation turns a complex-looking term into two simple, manageable components. This is the power of mathematical insight—transforming complexity into simplicity. ### The Telescoping Magic Now that we have our decomposed integrand, the integration becomes trivial. We evaluate the integral as follows: $$\int_{k}^{k+1} \left( \frac{1}{x} - \frac{1}{x+1} \right) dx = \left[ \ln|x| - \ln|x+1| \right]_{k}^{k+1}$$ Using the properties of logarithms, this simplifies to $\ln(\frac{x}{x+1})$. Applying the limits from $k$ to $k+1$, we obtain: $$\ln\left(\frac{k+1}{k+2}\right) - \ln\left(\frac{k}{k+1}\right)$$ Let us define a function $f(k) = \ln(\frac{k}{k+1})$. Our integral now takes the form $f(k+1) - f(k)$. When we sum this from $k=1$ to $98$, we get: $$[f(2) - f(1)] + [f(3) - f(2)] + \dots + [f(99) - f(98)]$$ This is a **telescoping series**. The terms cancel out like falling dominoes, leaving us with only $f(99) - f(1)$. ### Final Calculation The sum collapses into a single, elegant result: $$I = \ln\left(\frac{99}{100}\right) - \ln\left(\frac{1}{2}\right)$$ Simplifying the logarithmic expression: $$I = \ln\left( \frac{99/100}{1/2} \right) = \ln\left(\frac{99}{50}\right)$$ ### The Inequality Battle We have our value $I = \ln(\frac{99}{50})$. To compare this with other values, we use the standard inequality $\ln(1+x) 0$. Rewrite the argument as: $$\frac{99}{50} = 1 + \frac{49}{50}$$ Now, our expression is $\ln(1 + \frac{49}{50})$. By setting $x = \frac{49}{50}$, the inequality $\ln(1 + \frac{49}{50}) < \frac{49}{50}$ holds true. This confirms that **$I < \frac{49}{50}$**. Remember, every complex problem is just a sequence of simple steps; keep practicing to master these patterns.

JEE(ADVANCED)-201

LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: If $I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{d x}{x ( x + 1 )}$ , then

Q: Why did we use partial fractions instead of just integrating directly?

Great question! While you could try other methods, partial fractions are the most efficient way to break down a rational function into simpler components. It turns a complex product in the denominator into a simple difference, which is exactly what we need for a telescoping series.

Q: How do I know it's a telescoping series?

Look for the structure $f(k+1) - f(k)$. Whenever you see a sum where the terms are differences of the same function evaluated at consecutive points, it's a telescoping series. The intermediate terms will always cancel out, leaving only the first and last terms.

Q: How do I compare $\ln(\frac{99}{50})$ with $\frac{49}{50}$ without a calculator?

This is a classic JEE trick! We use the standard inequality $\ln(1+x) 0$. By rewriting $\frac{99}{50}$ as $1 + \frac{49}{50}$, we can directly apply this inequality with $x = \frac{49}{50}$ to prove the result.

Select Answer:

* Multiple Correct

Visualized Solution

$I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{d x}{x ( x + 1 )}$

We need to evaluate the given expression and check the inequalities.
$I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{d x}{x ( x + 1 )}$

$\int \frac{d x}{x ( x + 1 )}$

Let's first evaluate the indefinite integral.
Using partial fraction decomposition:
$\frac{1}{x ( x + 1 )} = \frac{1}{x} - \frac{1}{x + 1}$

$\int (\frac{1}{x} - \frac{1}{x + 1}) d x$

Now, integrate the decomposed expression:
$\int (\frac{1}{x} - \frac{1}{x + 1}) d x = ln ∣ x ∣ - ln ∣ x + 1∣$
Using $ln a - ln b = ln (\frac{a}{b})$ :
$= ln \frac{x}{x + 1}$

$[ln (\frac{x}{x + 1})]_{k}^{k + 1}$

Apply the limits from $k$ to $k + 1$ :
Upper limit: $ln (\frac{k + 1}{k + 2})$
Lower limit: $ln (\frac{k}{k + 1})$
Result: $ln (\frac{k + 1}{k + 2}) - ln (\frac{k}{k + 1})$

$I = \sum_{k = 1}^{98} [f (k + 1) - f (k)]$

Substitute this back into the original summation.
Let $f (k) = ln (\frac{k}{k + 1})$
The sum becomes: $\sum_{k = 1}^{98} [f (k + 1) - f (k)]$
This is a telescoping series!

$I = f (99) - f (1)$

Expand the sum:
$[f (2) - f (1)] + [f (3) - f (2)] + \dots + [f (99) - f (98)]$
All intermediate terms cancel out!
$I = f (99) - f (1)$

$I = ln (\frac{99}{50})$

Substitute the values back:
$f (99) = ln (\frac{99}{100})$ and $f (1) = ln (\frac{1}{2})$
$I = ln (\frac{99}{100}) - ln (\frac{1}{2})$
$I = ln (\frac{99}{100} \times 2) = ln (\frac{99}{50})$

$I < ln 99$

We need to check options [A] and [B].
Compare $I = ln (\frac{99}{50})$ with $ln 99$ .
Since $\frac{99}{50} < 99$ and $ln x$ is strictly increasing:
$ln (\frac{99}{50}) < ln (99)$
Option [B] is correct.

\text{Standard Inequality: } \ln(1+x) < x

Now check options [C] and [D]: compare $I$ with $\frac{49}{50}$ .
How do we compare a logarithm with an algebraic fraction?
We use the standard inequality: $ln (1 + x) < x$ for $x > 0$ .
Let's visualize this on the graph.

$ln (1 + \frac{49}{50}) < \frac{49}{50}$

Rewrite the argument: $\frac{99}{50} = 1 + \frac{49}{50}$
So, $I = ln (1 + \frac{49}{50})$
Apply the inequality with $x = \frac{49}{50}$ :
$ln (1 + \frac{49}{50}) < \frac{49}{50}$
Option [C] is correct.

\text{Final Conclusion}

Key Takeaways:
- Partial fraction decomposition simplifies rational integrands.
- Telescoping series drastically reduce massive summations.
- Standard inequalities like $ln (1 + x) < x$ are crucial for comparing logs with rational numbers.
Correct Options: [B] and [C]

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

The expression provided is:

I = k = 1 \sum 98 \int_{k}^{k + 1} \frac{d x}{x ( x + 1 )}

The key to solving this is to decompose the rational function

\frac{1}{x ( x + 1 )}

. Whenever you see a product in the denominator, your first instinct should be partial fraction decomposition.

We can rewrite the integrand as:

\frac{1}{x ( x + 1 )} = \frac{1}{x} - \frac{1}{x + 1}

This transformation turns a complex-looking term into two simple, manageable components. This is the power of mathematical insight—transforming complexity into simplicity.

The Telescoping Magic

Now that we have our decomposed integrand, the integration becomes trivial. We evaluate the integral as follows:

\int_{k}^{k + 1} (\frac{1}{x} - \frac{1}{x + 1}) d x = [ln ∣ x ∣ - ln ∣ x + 1∣]_{k}^{k + 1}

Using the properties of logarithms, this simplifies to

ln (\frac{x}{x + 1})

. Applying the limits from

k

k + 1

, we obtain:

ln (\frac{k + 1}{k + 2}) - ln (\frac{k}{k + 1})

Let us define a function

f (k) = ln (\frac{k}{k + 1})

. Our integral now takes the form

f (k + 1) - f (k)

When we sum this from

k = 1

98

, we get:

[f (2) - f (1)] + [f (3) - f (2)] + \dots + [f (99) - f (98)]

This is a telescoping series. The terms cancel out like falling dominoes, leaving us with only

f (99) - f (1)

Final Calculation

The sum collapses into a single, elegant result:

I = ln (\frac{99}{100}) - ln (\frac{1}{2})

Simplifying the logarithmic expression:

I = ln (\frac{99/100}{1/2}) = ln (\frac{99}{50})

The Inequality Battle

We have our value

I = ln (\frac{99}{50})

. To compare this with other values, we use the standard inequality

ln (1 + x) < x

for

x > 0

Rewrite the argument as:

\frac{99}{50} = 1 + \frac{49}{50}

Now, our expression is

ln (1 + \frac{49}{50})

. By setting

x = \frac{49}{50}

, the inequality

ln (1 + \frac{49}{50}) < \frac{49}{50}

holds true.

This confirms that $I < \frac{49}{50}$ . Remember, every complex problem is just a sequence of simple steps; keep practicing to master these patterns.

Common Traps & FAQs

Why did we use partial fractions instead of just integrating directly?

Great question! While you could try other methods, partial fractions are the most efficient way to break down a rational function into simpler components. It turns a complex product in the denominator into a simple difference, which is exactly what we need for a telescoping series.

How do I know it's a telescoping series?

Look for the structure

f (k + 1) - f (k)

. Whenever you see a sum where the terms are differences of the same function evaluated at consecutive points, it's a telescoping series. The intermediate terms will always cancel out, leaving only the first and last terms.

How do I compare

ln (\frac{99}{50})

with

\frac{49}{50}

without a calculator?

This is a classic JEE trick! We use the standard inequality

ln (1 + x) < x

for

x > 0

. By rewriting

\frac{99}{50}

1 + \frac{49}{50}

, we can directly apply this inequality with

x = \frac{49}{50}

to prove the result.

Animated Solution for Mathematics - Definite Integration: If $I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{d x}{x ( x + 1 )}$ , then

Select Answer:

Visualized Solution

$I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{d x}{x ( x + 1 )}$

$\int \frac{d x}{x ( x + 1 )}$

$\int (\frac{1}{x} - \frac{1}{x + 1}) d x$

$[ln (\frac{x}{x + 1})]_{k}^{k + 1}$

$I = \sum_{k = 1}^{98} [f (k + 1) - f (k)]$

$I = f (99) - f (1)$

$I = ln (\frac{99}{50})$

$I < ln 99$

\text{Standard Inequality: } \ln(1+x) < x

$ln (1 + \frac{49}{50}) < \frac{49}{50}$

\text{Final Conclusion}

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Telescoping Magic

Final Calculation

The Inequality Battle

Similar Questions

Let $α \in (0, 1)$ and $β = lo g_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3} + ... + \frac{x ^{n}}{n}, x \in (0, 1)$ . Then the integral $\int_{0}^{α} \frac{t ^{50}}{1 - t} d t$ is equal to

Consider the integral $I = \int_{0}^{10} \frac{[ x ] e ^{[x]}}{e ^{x - 1}} d x$ where $[x]$ denotes the greatest integer less than or equal to $x$ . Then the value of $I$ is equal to:

If $I_{n} = \int_{- π}^{π} \frac{s i n n x}{( 1 + π ^{x} ) s i n x} d x, n = 0, 1, 2, ..., t h e n$

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A value of $α$ such that $\int_{α}^{α + 1} \frac{d x}{( x + α ) ( x + α + 1 )} = lo g_{e} (9/8)$ is :

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The value of the integral $\int_{e^{- 1}}^{e^{2}} ∣ \frac{l o g _{e} x}{x} ∣ d x$ is

The greatest integer less than or equal to $\int_{1}^{2} lo g_{2} (x^{3} + 1) d x + \int_{1}^{l o g_{2} 9} (2^{x} - 1)^{1/3} d x$ is ______.

Let $a$ be a positive real number such that $\int_{0}^{a} e^{x - [x]} d x = 10 e - 9$ where $[x]$ is the greatest integer less than or equal to $x$ . Then $a$ is equal to :

The value of the integral $\int_{- 1}^{2} lo g_{e} (x + x^{2} + 1) d x$ is

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If I=∑k=198​∫kk+1​x(x+1)dx​, then

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=∑k=198​∫kk+1​x(x+1)dx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫x(x+1)dx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫(x1​−x+11​)dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [ln(x+1x​)]kk+1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=∑k=198​[f(k+1)−f(k)]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=f(99)−f(1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=ln(5099​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I<ln99

\text{Standard Inequality: } \ln(1+x) < x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(1+5049​)<5049​

\text{Final Conclusion}

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Telescoping Magic

Final Calculation

The Inequality Battle

Similar Questions

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Animated Solution for Mathematics - Definite Integration: If $I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{d x}{x ( x + 1 )}$ , then

$I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{d x}{x ( x + 1 )}$

$\int \frac{d x}{x ( x + 1 )}$

$\int (\frac{1}{x} - \frac{1}{x + 1}) d x$

$[ln (\frac{x}{x + 1})]_{k}^{k + 1}$

$I = \sum_{k = 1}^{98} [f (k + 1) - f (k)]$

$I = f (99) - f (1)$

$I = ln (\frac{99}{50})$

$I < ln 99$

$ln (1 + \frac{49}{50}) < \frac{49}{50}$

Let $α \in (0, 1)$ and $β = lo g_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3} + ... + \frac{x ^{n}}{n}, x \in (0, 1)$ . Then the integral $\int_{0}^{α} \frac{t ^{50}}{1 - t} d t$ is equal to

Consider the integral $I = \int_{0}^{10} \frac{[ x ] e ^{[x]}}{e ^{x - 1}} d x$ where $[x]$ denotes the greatest integer less than or equal to $x$ . Then the value of $I$ is equal to:

If $I_{n} = \int_{- π}^{π} \frac{s i n n x}{( 1 + π ^{x} ) s i n x} d x, n = 0, 1, 2, ..., t h e n$

Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}, x \in R - {- 1}, n \in N, n > 2$ . If $f_{n} (x) = (f \circ f \circ \dots upto n times) (x)$ , then $lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f_{n} (x)) d x$ is equal to

A value of $α$ such that $\int_{α}^{α + 1} \frac{d x}{( x + α ) ( x + α + 1 )} = lo g_{e} (9/8)$ is :

For any real number $x$ , let $[x]$ denote the largest integer less than or equal to $x$ . If $I = \int_{0}^{10} [\frac{10 x}{x + 1}] d x$ , then the value of $9 I$ is ___.

The value of the integral $\int_{e^{- 1}}^{e^{2}} ∣ \frac{l o g _{e} x}{x} ∣ d x$ is

The greatest integer less than or equal to $\int_{1}^{2} lo g_{2} (x^{3} + 1) d x + \int_{1}^{l o g_{2} 9} (2^{x} - 1)^{1/3} d x$ is ______.

Let $a$ be a positive real number such that $\int_{0}^{a} e^{x - [x]} d x = 10 e - 9$ where $[x]$ is the greatest integer less than or equal to $x$ . Then $a$ is equal to :

The value of the integral $\int_{- 1}^{2} lo g_{e} (x + x^{2} + 1) d x$ is