Question:

Let $\alpha \in(0,1)$ and $\beta=\log _e(1-\alpha)$ Let $P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\ldots+\frac{x^n}{n}, x \in(0,1)$ Then the integral $\int\limits_0^\alpha \frac{t^{50}}{1-t} d t$ is equal to

Updated On: Oct 13, 2024
  • $P_{50}(\alpha)-\beta$
  • $-\left(\beta+P_{50}(\alpha)\right)$
  • $\beta+P_{50}(a)$
  • $\beta-P_{50}(\alpha)$
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The Correct Option is B

Solution and Explanation

The correct answer is (B) : $-\left(\beta+P_{50}(\alpha)\right)$



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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives