Find an anti derivative (or integral) of the following function by the method of inspection: (ax + b)2
Solution and Explanation
The anti derivative of (ax + b)2 is the function of x whose derivative is (ax + b)2 .
It is known that,
\(\frac{d}{dx}\)(ax + b)3= 3a (ax + b)2
⇒ (ax + b)2 = \(\frac{1}{3a}\frac{d}{dx}\)(ax + b)3
∴ (ax + b)2 = \(\frac{d}{dx}\bigg(\frac{1}{3a}(ax+b)^3\bigg)\)
Therefore, the anti derivative of (ax + b)2 is \(\frac{1}{3a}\)(ax + b)3 .
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Questions Asked in CBSE CLASS XII exam
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What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
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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.