The integral \(∫_0^ 1 \frac{ 1}{7^{[\frac{1}{z}}]}dx\) where [⋅] denotes the greatest integer function, is equal to
- \(1+6log_e(\frac{6}{7})\)
- \(1-6log_e(\frac{6}{7})\)
- \(log_e(\frac{7}{6})\)
- \(1-7log_e(\frac{6}{7})\)
The Correct Option is A
Solution and Explanation
\(∫_0^ 1 \frac{ 1}{7^{[\frac{1}{z}}]}dx\), let \(\frac{1}{x}=t\)
\(\frac{−1}{x^2}dx=dt\)
=\(∫_∞ ^1 \frac{1}{−t^27[t]}dt=∫_1 ^∞ \frac{1}{t^27[t]}dt\)
=\(∫_1^ 2 \frac{1}{7t^2}dt+∫_2 ^3 \frac{1}{7^2t^2}dt+…\)
=\(\frac{1}{7}\bigg[−\frac{1}{t}\bigg]_1^ 2+\frac{1}{7 ^2}\bigg[\frac{−1}{t}\bigg]_2^ 3+\frac{1}{7 ^3}\bigg[\frac{−1}{t}\bigg]_2^ 3+…\)
=\(∑_{n=1} ^∞\frac{1}{7_n}\bigg(\frac{1}{n}−\frac{1}{n+1}\bigg)\)
=\(∑_{n=1} ^∞\frac{(\frac{1}{7})^n}{n}−7∑_{n=1} ^∞\frac{(\frac{1}{7})^{n+1}}{n+1}\)
=\(−log\bigg(1−\frac{1}{7}\bigg)+7log\bigg(1−\frac{1}{7}\bigg)+1\)
=\(1+6\log\frac{6}{7}\)
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.