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Estimated time: 10 minutes
CBSE: Class 12
Introduction
Definite integrals help in finding the accumulated value of a quantity over an interval. The properties of definite integrals are especially useful because they simplify lengthy expressions and reduce calculation time in board and entrance-exam questions.
These properties are based on symmetry, change of limits, splitting of intervals, and simple substitutions.
CBSE: Class 12
Properties of Definite Integrals
| Property | Formula | Meaning |
|---|---|---|
| \[P_0\] : Change of Variable |
\[\int_{a}^{b} f(x) dx = \int_{a}^{b} f(t) dt\]
|
Changing the variable of integration does not change the value of the integral. |
| \[P_1\] : Reversing Limits |
\[\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx\]
|
Swapping the upper and lower limits changes the sign of the integral. |
| \[P_2\] : Splitting the Interval |
\[\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx\]
|
Useful for piecewise functions or modulus (absolute value) functions. |
| \[P_3\] : The "King's Rule" |
\[\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx\]
|
Symmetry on [a,b] |
| \[P_4\] : Special Case of $P_3$ |
\[\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx\]
|
Applied when the lower limit is zero. |
| \[P_5\] : Halving the Upper Limit (Addition) |
\[\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(2a - x) dx\]
|
Halving the Upper Limit |
| \[P_6\] : Halving the Upper Limit (Conditional)) |
\[\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx \quad \text{if } f(2a - x) = f(x)\]
\[\int_{0}^{2a} f(x) dx = 0 \quad \text{if } f(2a - x) = -f(x)\]
|
Derived from \[P_5\], this simplifies integrals based on the function's behavior at (2a - x). |
| \[P_7\] : Even and Odd Functions (Symmetric Limits) |
(i) Even Function: If f(-x) = f(x) \[\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\]
(ii) Odd Function: If f(-x) = -f(x) \[\int_{-a}^{a} f(x) dx = 0\]
|
Used when limits are from -a to a. |
CBSE: Class 12
Example 1
Evaluate \[\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x dx\]
Solution:
We observe that \[\sin^2 x\] is an even function. Therefore, by \[P_7\] (i), we get
\[\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^2 x dx\]
\[= 2 \int_{0}^{\frac{\pi}{4}} \frac{(1 - \cos 2x)}{2} dx = \int_{0}^{\frac{\pi}{4}} (1 - \cos 2x) dx\]
\[= \left[ x - \frac{1}{2} \sin 2x \right]_{0}^{\frac{\pi}{4}} = \left( \frac{\pi}{4} - \frac{1}{2} \sin \frac{\pi}{2} \right) - 0 = \frac{\pi}{4} - \frac{1}{2}\]
CBSE: Class 12
Example 2
Evaluate \[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}}\]
Solution: Let \[I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x} dx}{\sqrt{\cos x} + \sqrt{\sin x}} \quad \dots (1)\]
Then, by \[P_3\]
\[I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}}{\sqrt{\cos \left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)} + \sqrt{\sin \left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}} dx\]
\[= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \quad \dots (2)\]
Adding (1) and (2), we get
\[2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} dx = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\]
Hence
\[I = \frac{\pi}{12}\]
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