Question:

The anti derivative of \(\bigg(\sqrt x+\frac{1}{\sqrt x}\bigg)\) equals

Updated On: Sep 15, 2023
  • \(\frac{1}{3}x^{\frac{1}{3}}+2x^{\frac{1}{2}}+C\)

  • \(\frac{2}{3}x^{\frac{2}{3}}+\frac{1}{2}x^{\frac{1}{2}}+C\)

  • \(\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\)

  • \(\frac{2}{3}x^{\frac{3}{2}}+\frac{1}{2}x^{\frac{1}{2}}+C\)

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The Correct Option is C

Solution and Explanation

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

\(\int x^{\frac{1}{2}}dx+\int x^{\frac{1}{2}}dx\)

\(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C\)

=\(\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\)

Hence, the correct Answer is C.

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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:

  1. The problem of getting a function if its derivative is given.
  2. The problem of getting the area bounded by the graph of a function under given situations.