it is convenient to write it in the equivalent differential form $dy=f(x)dx$
The operation of finding all solutions of this equation is called antidifferetiation [0r indefinite integration] adn is denoted by an integral sign $\int$.
The general solution is denoted by $y=\int f(x)dx=F(x)+C$
\begin{split}
f(x)&= integrand\\
d(x)&= Variable\ of\ integration\\
F(x)&=An\ antiderivative\ of\ f(x)\\
C&=Constant\ of\ integration
\end{split}
The expression $\int f(x)dx$ is read as the antiderivative of $f$ with respect to $x$. So, the differential $dx$ serves to identify $x$ as the variable of integration. The term indefinite integral is a synonym for antiderivative.
Basic Integration Rules
The inverse nature of integration and differentiation can be verified by substituting $F'(x)$ for $f(x)$ in the indefinite integration definotion to obtain;$\int F'(x)dx=F(x)+c$
Moreover, if $\int f(x)dx=F(x)+C$, then
$\dfrac{d}{dx}\left [ \int f(x)dx) \right ]=f(x)$
- $\int x^{n}dx=\dfrac{1}{n+1}x^{n}+C,\ n\neq -1$
- $\int kf(x)dx=k\int f(x)dx$
- $\int [f(x) \pm g(x)]dx=\int f(x)dx \pm \int g(x)dx$
- $\int dx= x + C$
- $\int a^{x} dx= \left (\dfrac{1}{ln\ a} \right )a^{x} + C$
- $\int e^{x} dx= e^{x} + C$
- $\int sin\ x\ dx= -cos\ x + C$
- $\int sin\ u\ dx= -\dfrac{1}{u'}cos\ u + C$
- $\int cos\ x\ dx= sin\ x + C$
- $\int cos\ u\ dx= \dfrac{1}{u'}sin\ u + C$
- $\int tan\ x\ dx= -ln|cos\ x| + C$
- $\int cot\ x\ dx= ln|sin\ x| + C$
- $\int sec\ x\ dx= ln|sec\ x+tan\ x| + C$
- $\int csc\ x\ dx= -ln|csc\ x+cot\ x| + C$
- $\int sec^{2}\ x\ dx= tan\ x + C$
- $\int sec^{2}\ u\ dx=\dfrac{1}{u'} tan\ u + C$
- $\int csc^{2}\ x\ dx= -cot\ x + C$
- $\int csc^{2}\ u\ dx= -\dfrac{1}{u'}cot\ u + C$
- $\int sec\ x\ tan\ x\ dx= sec\ x + C$
- $\int csc\ x\ cot\ x\ dx= -csc\ x + C$
- $\int \dfrac{dx}{\sqrt{a^{2}-x^{2}}}=arcsin\dfrac{u}{a}+C$
- $\int \dfrac{dx}{a^{2}+x^{2}}=\dfrac{1}{a}arctan\dfrac{u}{a}+C$
- $\int \dfrac{dx}{x\sqrt{x^{2}-a^{2}}}=\dfrac{1}{a}arcsec\dfrac{|u|}{a}+C$
Partial Integration
$\int u\ dv=u \cdot v-\int v\ du$The Fundamental Theorem Of Calculus
If a function $f$ is continuous on the closed interval $[a,b]$ and $F$ is an antiderivative of $f$ on the interval $[a,b]$, then$\int_{a}^{b}f(x)dx=F(b)-F(a)$
Daftar bacaan:
- Calculus Ninth Edition by Varberg, Purcel and Rigdon
- Calculus Ninth Edition by Larson, and Edwards
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