Introduction
Definite Integration is an essential topic in calculus and is crucial for solving many problems in the JEE Main Mathematics syllabus. It involves finding the exact area under a curve defined by a function over a specified interval. This study note will break down the concepts of definite integration, covering the fundamental theorem of calculus, properties of definite integrals, and techniques for evaluating them.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the concept of differentiation with integration and is divided into two parts:
First Part
The first part states that if $f$ is a continuous real-valued function defined on a closed interval $[a, b]$, and $F$ is the antiderivative of $f$ on $[a, b]$, then:
$$ \int_a^b f(x) , dx = F(b) - F(a) $$
NoteThe antiderivative $F(x)$ is a function such that $F'(x) = f(x)$.
Second Part
The second part of the theorem states that if $F$ is an antiderivative of $f$ on an interval $I$, then for every $x$ in $I$:
$$ \frac{d}{dx} \left( \int_a^x f(t) , dt \right) = f(x) $$
Properties of Definite Integrals
Definite integrals have several important properties that simplify calculations and help in solving complex problems. Here are some key properties:
Linearity
If $a$ and $b$ are constants, then:
$$ \int_a^b [cf(x) + dg(x)] , dx = c \int_a^b f(x) , dx + d \int_a^b g(x) , dx $$
Additivity
If $a
< c < b$, then:
$$ \int_a^b f(x) , dx = \int_a^c f(x) , dx + \int_c^b f(x) , dx $$
Symmetry
If $f(x)$ is an even function, i.e., $f(-x) = f(x)$, then:
$$ \int_{-a}^a f(x) , dx = 2 \int_0^a f(x) , dx $$
If $f(x)$ is an odd function, i.e., $f(-x) = -f(x)$, then:
$$ \int_{-a}^a f(x) , dx = 0 $$
Mean Value Theorem for Integrals
If $f$ is continuous on $[a, b]$, then there exists a number $c$ in $(a, b)$ such that:
$$ \int_a^b f(x) , dx = f(c)(b - a) $$
TipUse symmetry properties to simplify integrals involving even or odd functions.
Techniques for Evaluating Definite Integrals
Substitution
If $u = g(x)$ is a differentiable function whose range is an interval $I$, and $f$ is a function continuous on $I$, then:
$$ \int_a^b f(g(x))g'(x) , dx = \int_{g(a)}^{g(b)} f(u) , du $$
ExampleEvaluate the integral $\int_0^1 2x \cos(x^2) , dx$.
Let $u = x^2$, then $du = 2x , dx$. When $x = 0$, $u = 0$. When $x = 1$, $u = 1$.
Thus, the integral becomes:
$$ \int_0^1 \cos(x^2) \cdot 2x , dx = \int_0^1 \cos(u) , du = \sin(u) \Big|_0^1 = \sin(1) - \sin(0) = \sin(1) $$
Integration by Parts
If $u = f(x)$ and $dv = g(x) , dx$, then:
$$ \int_a^b u , dv = \left[ uv \right]_a^b - \int_a^b v , du $$
ExampleEvaluate $\int_0^1 x e^x , dx$.
Let $u = x$ and $dv = e^x , dx$. Then $du = dx$ and $v = e^x$.
$$ \int_0^1 x e^x , dx = \left[ x e^x \right]_0^1 - \int_0^1 e^x , dx = \left[ x e^x \right]_0^1 - \left[ e^x \right]_0^1 $$
$$ = (1 \cdot e^1 - 0 \cdot e^0) - (e^1 - e^0) = e - (e - 1) = 1 $$
Partial Fractions
For rational functions, express the integrand as a sum of simpler fractions and then integrate.
ExampleEvaluate $\int_0^1 \frac{2}{x^2 - 1} , dx$.
First, decompose into partial fractions:
$$ \frac{2}{x^2 - 1} = \frac{2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} $$
Solving, we find $A = 1$ and $B = -1$:
$$ \int_0^1 \left( \frac{1}{x-1} - \frac{1}{x+1} \right) , dx $$
Integrate each term separately:
$$ \left[ \ln|x-1| - \ln|x+1| \right]_0^1 = \ln|0| - \ln|2| - (\ln|-1| - \ln|1|) = -\ln(2) + \ln(1) = -\ln(2) $$
Common MistakeA common mistake is to forget to change the limits of integration when using substitution. Always remember to adjust the limits according to the new variable.
Applications of Definite Integrals
Area Under a Curve
The area under the curve $y = f(x)$ from $x = a$ to $x = b$ is given by:
$$ \text{Area} = \int_a^b f(x) , dx $$
Area Between Two Curves
If $f(x) \geq g(x)$ for all $x$ in $[a, b]$, the area between the curves $y = f(x)$ and $y = g(x)$ is:
$$ \text{Area} = \int_a^b [f(x) - g(x)] , dx $$
ExampleFind the area between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$.
$$ \text{Area} = \int_0^1 (x - x^2) , dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{1}{6} $$
Volume of Solids of Revolution
Using the disk method, the volume of a solid of revolution formed by rotating $y = f(x)$ about the x-axis from $x = a$ to $x = b$ is:
$$ \text{Volume} = \pi \int_a^b [f(x)]^2 , dx $$
TipWhen dealing with volumes of revolution, visualize the shape formed by the rotation to set up the integral correctly.
Conclusion
Definite Integration is a powerful tool in calculus with numerous applications. Understanding the fundamental theorem of calculus, properties of definite integrals, and various techniques for evaluating them is crucial for solving problems in the JEE Main Mathematics syllabus. Practice these concepts with a variety of problems to gain proficiency and confidence.
