Find the following integral: \(\int (2x^2-3sinx+5\sqrt x)dx\)
Solution and Explanation
\(\int (2x^2-3sinx+5\sqrt x)dx\)
= \(2 \int x^2dx-3 \int \sin xdx+5 \int x^{\frac{1}{2}}dx\)
= \(\frac{2x^3}{3}-3(- \cos x)+5 \bigg(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\bigg)+C\)
=\(\frac{2}{3}x^3+3\cos x+ \frac{10}{3}x^{\frac{3}{2}}+C\)
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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.