Revision: Calculus >> Integrals Maths Commerce (English Medium) Class 12 CBSE

If f(x) is a continuous function defined on an interval [a, b] and if Φ(x) is the antiderivative of f(x), i.e., \[\frac{d}{dx}[\phi(x)]=f(x)\] then the definite integral of f(x) over [a, b] denoted by \[\int_{a}^{b}f(x)dx\]  is defined as

\[\int_{a}^{b}f(x)dx=
\begin{bmatrix}
\phi\left(x\right)
\end{bmatrix}_{a}^{b}=\phi\left(b\right)-\phi\left(a\right)\]

\[\mathrm{If~}\frac{d}{dx}[F(x)]=f(x),\mathrm{~then~}\int f(x)dx=F(x)\]

Integration is the inverse process of differentiation.

\[\int f(x)dx=F(x)+c\]

• The arbitrary constant 'c' is called the constant of integration.
• F(x) + c is called the indefinite integral.

1. \[\int[f(x)]^nf^{\prime}(x)dx=\frac{[f(x)]^{n+1}}{n+1}+c\quad(n\neq-1)\]

2. \[\int\frac{f^{\prime}(x)}{f(x)}dx=\log|f(x)|+c\]

\[\int e^x\left[\left.f(x)+f^{\prime}(x)\right.\right]dx=e^xf(x)+c\]

If, u = f(x) ⇒ \[\frac{du}{dx}=f^{\prime}(x)\]

then \[\int[f(x)]^nf^{\prime}(x)dx=\frac{[f(x)]^{n+1}}{n+1}+c\quad(n\neq-1)\]

Linear Substitution Rule:

If $b$, then

\[\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+c\quad(n\neq-1)\]

Statement:

If f(x) and g(x) are any two differentiable functions of x and G(x) is the antiderivative of g(x), i.e., \[G(x)=\int g(x)dx\]. Then 

\[\int f(x)g(x)dx=f(x)G(x)-\int f^{\prime}(x)G(x)dx\]

(A) Non-repeated linear factors

\[\frac{Ax+B}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}\]

(B) Repeated linear factor

\[\frac{Ax+B}{(x-a)^n}=\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_n}{(x-a)^n}\]

(C) Quadratic factor (not factorisable)

\[\frac{Ax+B}{ax^2+bx+c}\]

1.\[\int\sqrt{(a^{2}-x^{2})}dx=\frac{1}{2}x\sqrt{(a^{2}-x^{2})}+\frac{1}{2}a^{2}\sin^{-1}\left(\frac{x}{a}\right)+c\]

2. \[\int\left(\sqrt{a^{2}+x^{2}}\right)dx=\frac{1}{2}x\sqrt{(a^{2}+x^{2})}+\frac{1}{2}a^{2}\log|x+\sqrt{(a^{2}+x^{2})}|+c\]

Let I = `int sqrt(a^2 - x^2) dx`

= `int sqrt(a^2 - x^2)*1 dx`

= `sqrt(a^2 - x^2)* int 1 dx - int [d/dx (sqrt(a^2 - x^2))* int 1 dx]dx`

= `sqrt(a^2 - x^2)*x - int [1/(2sqrt(a^2 - x^2))*d/dx (a^2 - x^2)*x]dx`

= `sqrt(a^2 - x^2)*x - int 1/(2sqrt(a^2 - x^2))(0 - 2x)*x  dx`

= `sqrt(a^2 - x^2)*x - int (-x)/sqrt(a^2 - x^2)*x  dx`

= `xsqrt(a^2 - x^2) - int (a^2 - x^2 - a^2)/sqrt(a^2 - x^2)dx`

= `xsqrt(a^2 - x^2) - int sqrt(a^2 - x^2)dx + a^2 int dx/sqrt(a^2 - x^2)`

= `xsqrt(a^2 - x^2) - I + a^2sin^-1(x/a) + c_1`

∴ 2I = `xsqrt(a^2 - x^2) + a^2sin^-1(x/a) + c_1`

∴ I = `x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1(x/a) + c_1/2`

∴ `int sqrt(a^2 - x^2)dx = x/2 sqrt(a^2 - x^2) + a^2/2sin^-1(x/a) + c`, where `c = c_1/2`.

Theorem 1:

 Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function. Then A′(x) = f (x), for all x ∈ [a, b]

Theorem 2:

Let f  be a continuous function defined on the closed interval [a, b], and F be an antiderivative of f. Then  \[\int_a^bf(x)dx=\left[\mathbf{F}(x)\right]_a^b=\mathbf{F}(b)-\mathbf{F}(a)\]

For choosing the first function:

L I A T E

• Logarithmic

• Inverse trigonometric

• Algebraic

• Trigonometric

• Exponential

1.\[\frac{d}{dx}{\left[\int f(x)dx\right]}=f(x)\]

2. \[\int cf(x)dx=c\int f(x)dx\]

3. \[\int(u+v-w)dx=\int udx+\int vdx-\int wdx\]