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Definite integration formulas

Definite integration involves finding the integral of a function over a specific interval. This process yields a number that represents the net area under the curve of the function between the two endpoints of the interval. Here are some fundamental formulas and properties related to definite integrals:

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration and consists of two parts:

1. First Part: If ( f ) is continuous on ([a, b]) and ( F ) is an antiderivative of ( f ) on ([a, b]), then:
$\int_a^b f(x) \, dx = F(b) – F(a)$
2. Second Part: If ( f ) is a continuous function on ([a, b]), then the function ( g ) defined by:
$g(x) = \int_a^x f(t) \, dt$
is continuous on ([a, b]), differentiable on ((a, b)), and ( g'(x) = f(x) ).

Properties of Definite Integrals

(I) $\int_{a}^{b} f(x) dx= \int_{a}^{b} f(t) dt$

(II) $\int_{a}^{b} f(x) dx=- \int_{b}^{a} f(x) dx$

(III) $\int_{a}^{b} f(x) dx= \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$

where a<c <b

(IV) $\int_{a}^{b} f(x) dx=\int_{a}^{b} f((a+b-x)) dx$

(V) $\int_{0}^{a} f(x) dx=\int_{0}^{a} f((a-x)) dx$

(VI) $\int_{0}^{2a} f(x) dx=\int_{0}^{a} f((x)) dx + \int_{0}^{a} f((2a-x)) dx$

(VII) $\int_{0}^{2a} f(x) dx= \begin{cases} 2 \int_{0}^{a} f(x) dx & , f(2a-x) =f(x) \\ 0 &, f(2a-x) =-f(x) \end{cases}$

(VIII) $\int_{-a}^{a} f(x) dx= \begin{cases} 2 \int_{0}^{a} f(x) dx & , f(x) =f(-x) \\ 0 &, f(x) =-f(x) \end{cases}$

(IX) $f(x) \geq 0$, then $\int_{a}^{b} f(x) dx \geq 0$

(X) $f(x) \geq g(x)$, then $\int_{a}^{b} f(x) dx \geq \int_{a}^{b} g(x) dx$

(IX) $|\int_{a}^{b} f(x) dx| \leq \int_{a}^{b} |f(x)| dx$

Other formula’s

For any real number $n \neq -1$,
$$\int_{a}^{b} x^n \, dx = \frac{b^{n+1} – a^{n+1}}{n+1}$$

$$\int_{a}^{b} e^x \, dx = e^b – e^a$$

$$\int_{1}^{b} \frac{1}{x} \, dx = \ln(b)$$

$$\int_{a}^{b} \sin(x) \, dx = -\cos(b) + \cos(a)$$
$$\int_{a}^{b} \cos(x) \, dx = \sin(b) – \sin(a)$$
$$\int_{a}^{b} \tan(x) \, dx = -\ln|\cos(b)| + \ln|\cos(a)|$$

A special case using polar coordinates:
$$\int_{0}^{2\pi} \frac{1}{2} r^2 \, d\theta = \pi r^2$$

$$\int_{a}^{\infty} \frac{1}{a^2 + x^2} \, dx = \frac {\pi}{2a}$$

$$\int_{a}^{\infty} \frac{1}{\sqrt {a^2 – x^2}} \, dx = \frac {\pi}{2}$$

$$\int_{a}^{\infty} \sqrt {a^2 – x^2} \, dx = \frac {\pi a^2}{4}$$

Example

Calculate the definite integral of ( f(x) = x^3 ) from ( x = 1 ) to ( x = 2 ):

$\int_1^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_1^2 = \frac{2^4}{4} – \frac{1^4}{4} = \frac{16}{4} – \frac{1}{4} = 4 – 0.25 = 3.75$

We hope that these Definite integration formulas will help in your preparation.

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