Maharashtra Board Class 12 Maths Part 2 Chapter 4 Definite Integration PDF Download

4. Definite Integration

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Definite integral as limit of sum.

Fundamental theorem of integral calculus.

Methods of evaluation and properties of definite integral.

4.1 Definite Integral As Limit Of Sum

In the last chapter, we studied various methods of finding the primitives or indefinite integrals of given function. We shall now interpret the definite integrals denoted by \(\int_a^b f(x) dx\), read as the integral from a to b of the function f(x) with respect to x. Here a < b are real numbers and f(x) is defined on [a, b]. At present, we assume that f(x) ≥ 0 on [a, b] and f(x) is continuous.

\(\int_a^b f(x) dx\) is defined as the area of the region bounded by y = f(x), x-axis and the ordinates x = a and x = b. If g(x) is the primitive of f(x), then the area is g(b) − g(a).

The reason of the above definition will be clear from the figure 4.1, and the discussion that follows here. We are using the mean value theorem learnt earlier. Divide the interval [a, b] into n equal parts by

\(a = x_0 < x_1 < x_2 < \ldots < x_{n-1} < x_n = b\)

Draw the curve y = f(x) in [a, b] and divide the interval [a, b] into n equal parts by

\(a = x_0 < x_1 < x_2 < \ldots < x_{n-1} < x_n = b\)

Divide the region whose area is measured into their strips as above.

Note that, the area of each strip can be approximated by the area of a rectangle M_r M_{r+1} QP as shown in the figure 4.1, which is (x_{r+1} − x_r) × f(T_r) where T is a point on the curve y = f(x) between P and Q.

Teacher's Note

Definite integral means finding total area under a curve between two points. Think of it like measuring the total land between a river curve and a straight road from point A to point B.

Exam Trick

Remember: Definite integral = area under curve. The upper limit minus lower limit gives the answer. Just like finding marks from first exam to last exam.

Points to Remember

Definite integral has two limits: lower limit (a) and upper limit (b).
The answer is always a number, not a function.
We use the Fundamental Theorem to calculate it quickly.
The area is found by subtracting g(a) from g(b).

4.2 Fundamental Theorem Of Integral Calculus

Let f be the continuous function defined on [a, b] and if \(\int f(x) dx = g(x) + c\) then \(\int_a^b f(x) dx = [g(x) + c]_a^b = [(g(b) + c) − (g(a) + c)] = g(b) + c − g(a) − c = g(b) − g(a)\)

Thus \(\int_a^b f(x) dx = g(b) − g(a)\)

In \(\int_a^b f(x) dx\), a is called as a lower limit and b is called as an upper limit.

Now let us discuss some fundamental properties of definite integration. These properties are very useful in evaluation of the definite integral.

4.2.1

Property I

\(\int_a^a f(x) dx = 0\)

Let \(\int f(x) dx = g(x) + c\)

\(\int_a^a f(x) dx = [g(x) + c]_a^a = [(g(a) + c) − (g(a) + c)] = 0\)

Property II

\(\int_a^b f(x) dx = −\int_b^a f(x) dx\)

Let \(\int f(x) dx = g(x) + c\)

\(\int_a^b f(x) dx = [g(x) + c]_a^b = [(g(b) + c) − (g(a) + c)] = g(b) − g(a) = −[g(a) − g(b)] = −\int_b^a f(x) dx\)

Thus \(\int_a^b f(x) dx = −\int_b^a f(x) dx\)

Ex. \(\int_1^3 x dx = \left[\frac{x^2}{2}\right]_1^3 = \frac{9}{2} − \frac{1}{2} = 4\)

Ex. \(\int_3^1 x dx = \left[\frac{x^2}{2}\right]_3^1 = \frac{1}{2} − \frac{9}{2} = −4\)

Teacher's Note

Same limits mean zero area. If we measure the land from point A to point A itself, there is no distance, so no area.

Exam Trick

Remember: If upper and lower limits are the same, answer is always zero. If you swap the limits, the answer becomes negative.

Points to Remember

\(\int_a^a f(x) dx = 0\) always.
Swapping limits changes the sign of the answer.
Property III says the answer is independent of the variable name.
Property IV allows us to break the integral at any point in between.

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