The anti derivative of \(\bigg(\sqrt x+\frac{1}{\sqrt x}\bigg)\) equals
The anti derivative of \(\bigg(\sqrt x+\frac{1}{\sqrt x}\bigg)\) equals
\(\frac{1}{3}x^{\frac{1}{3}}+2x^{\frac{1}{2}}+C\)
\(\frac{2}{3}x^{\frac{2}{3}}+\frac{1}{2}x^{\frac{1}{2}}+C\)
\(\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\)
\(\frac{2}{3}x^{\frac{3}{2}}+\frac{1}{2}x^{\frac{1}{2}}+C\)
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:
- 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.