Definite Integration and Its Applications Class 11 Physics - Mathematical Tools | NEET 2024

Previously, we learnt about algebraic, trigonometric, exponential and logarithmic functions. In calculus, we discussed differentiation of a function, properties of derivatives, formulas and application of derivatives. We also have gone through the integration of function and some basic formulas to determine the integral of a given function. In this article, we shall discuss further properties of integrals, definite integrals and their applications.

Refer to previous articles on mathematical tools for physics Class 11:

• Functions
• Trigonometry
• Differential Calculus
• Application of Derivatives

Table of Contents:

• Properties of Integrals
• Definite Integration
• Area Under the Curve
• Integration by Substitution

Properties of Integrals

As we have already discussed, integrals are inverse of derivatives. If we take the integral of the derivative of a function, we get the original function. Now, we shall discuss the properties of integrals which are essential to find the integral various functions:

• ∫ a f(x) dx = a ∫ f(x) dx where a is a constant
• ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
• ∫ [af(x) ± bg(x)] dx = a ∫ f(x) dx ± b ∫ g(x) dx where a and b are constants.
• If ∫ f(x) dx = F(x) + C, then ∫ f(ax + b) dx = 1/a F(ax + b) + C

For example, we have to find the integration of e ^{5x + 2}. We use the last property

Now, ∫ e^x dx = e^x + c

Then ∫ e ^{5x + 2} dx = ⅕(e^{5x + 2}) + C

Definite Integration

Integrating within limits is known as definite integrals. Let f(x) be a function which is needed to integrate within the interval [a, b], then

Where F(x) is the anti-derivative of f(x).

a is the lower limit, and b is the upper limit.

Geometrically, the integration of f(x) within [a, b] is the area under the curve f(x) from a to b.

Let us take an example,

Area Under the Curve

The area under the curve f(x) from x = a to x = b is calculated by dividing the region from x = a to x = b into very small rectangular strips of width dx and then summing up all these strips.

Let us take an example to understand this. Suppose we have to calculate the area of a parabola represented by y = x² from x = 0 to x = 2.

The area under the parabola = Area under the curve y = x² from x = 0 to x = 2.

= 2³/3 – 0³/3 = 8/3 – 0 = 8/3 units.

Now, we shall learn about integration methods, in which we shall first discuss integration by substitution.

Integration by Substitution

Integration by substitution can also be regarded as the reverse of the chain rule for differentiation. Let us suppose that I = ∫ f(x) dx. Substitute x = g(t) such that dx = g’(t) dt, then the given integral I can be transformed into

I = ∫ f(x) dx = ∫ f(g(t)) g’(t)dt.

This method of substitution is applied when we directly cannot apply the formulas of integration. Let us take an example, and we have to find the integral of the function 2x/(1 + x²).

We see that the numerator of the function is the derivative of the denominator. Hence we can apply the method of substitution.

Let 1 + x² = u then 2x dx = du, we have

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