Learn indefinite integrals with examples. Solving indefinite integrals of trigonometric functions. Evaluating Definite Integral using FTC 2.
\(\int f(x)\;dx=F(x)\) means that \(F'(x)=f(x)\). We say that \( 'F'\) is the indefinite integral of \( 'f'\).
\(\int x^3\;dx=\dfrac {x^4}{4}+C\rightarrow\) each value of 'C' will give an anti-derivative.
\(\therefore \) \(\dfrac {x^4}{4}+1,\;\dfrac {x^4}{4}-2,\;\;;\dfrac {x^4}{4}-6\) are all anti-derivatives of \(x^3\).
\(\int \limits_a^b f(x)dx=\int f(x)dx\Big]_a^b\)
Consider
\(I=\int\;x^2\;dx=\underbrace {\dfrac {x^3}{3}+C}_{function \;of \;x}\)
whereas \(\int\limits_1^2\;x^2\;dx=\dfrac {x^3}{3} \Bigg]_1^2=\dfrac {8}{3}-\dfrac {1}{3}=\underbrace{\dfrac {7}{3}}_{value}\)
A \(-cos\,e^x+C\)
B \(\dfrac {x^2}{2}+C\)
C \(-sin\,e^x+C\)
D \(x^3+C\)
(1) \(\int c\,f(x)\;dx=c\int\,f(x)\;dx\)
(2) \(\int \Big(f(x)+g(x)\Big)\;dx=\int f(x)dx+\int g(x)\;dx\)
(3) \(\int x^n\;dx=\dfrac {x^n+1}{n+1}+C\) (for all constants \(n\))
(4) \(\int k\;dx=kx+C\) where \('k'\) is a constant
A \(x^7+x^5-\sqrt x+C\)
B \(2x^2-\sqrt x+7x\sqrt x+C\)
C \(4x^3+x\sqrt x-\dfrac {1}{x}+C\)
D \(\dfrac {2x^3}{3}+\dfrac {2}{3}x^{3/2}+2x^{-1/2}+C\)
\(\displaystyle\int_a^b\,f(x)\;dx=F(b)-F(a)\) where \(\displaystyle\int f(x)=F(x)+C\)
The integration constant 'C' has no role in the definite integral as it cancels out.
\(\displaystyle\int_a^b\,f(x)\;dx=F(x)+C\Bigg]_a^b\)
\(=\Big(F(b)+C\Big)-\Big(F(a)+C\Big)\)
\(=F(b)+C-F(a)-C\)
\(=F(b)-F(a)\)
A \(\dfrac {80}{3}\)
B \(14\)
C \(\dfrac {-17}{3}\)
D \(\dfrac {42}{5}\)
Rules of Indefinite Integrals:
(1) \(\int c\,f(x)\;dx=c\int\,f(x)\;dx\)
(2) \(\int \Big(f(x)+g(x)\Big)\;dx=\int f(x)dx+\int g(x)\;dx\)
(3) \(\int x^n\;dx=\dfrac {x^n+1}{n+1}+C\) (for all constants \(n\))
(4) \(\int k\;dx=kx+C\) where \('k'\) is a constant
(1) \(\displaystyle\int\,sin\,x\;dx=-cos\,x+C\)
(2) \(\displaystyle\int\,cos\,x\;dx=sin\,x+C\)
(3) \(\displaystyle\int\,sec^2\,x\;dx=tan\,x+C\)
(4) \(\displaystyle\int\,cosec^2\,x\;dx=-cot\,x+C\)
(5) \(\displaystyle\int\,sec\,x\;tan\,x\;dx=sec\,x+C\)
(6) \(\displaystyle\int\,cosec\,x\;cot\,x\;dx=-cosec\,x+C\)
All these can be verified by differentiating the right hand function.
A \(\dfrac {5}{4}\sqrt x+7x^{{7}/{2}}-sin\,x+C\)
B \(\dfrac {7}{8}\,cos^2x+5x^{3/2}+x-C\)
C \(\dfrac {2}{5}\,x^{5/2}+2x^{1/2}+\dfrac {1}{2}\,tan\,x+C\)
D \(8\,sin\,x-\dfrac {1}{x\sqrt x}+x^2+C\)
(1) \(\displaystyle\int\,sin\,x\;dx=-cos\,x+C\)
(2) \(\displaystyle\int\,cos\,x\;dx=sin\,x+C\)
(3) \(\displaystyle\int\,sec^2\,x\;dx=tan\,x+C\)
(4) \(\displaystyle\int\,cosec^2\,x\;dx=-cot\,x+C\)
(5) \(\displaystyle\int\,sec\,x\;tan\,x\;dx=sec\,x+C\)
(6) \(\displaystyle\int\,cosec\,x\;cot\,x\;dx=-cosec\,x+C\)
All these can be verified by differentiating the right hand function.
A \(2(tan\,x-x)+C\)
B \(2\,cos\,x-\dfrac {1}{x}+C\)
C \(2\,sec\,x+x+C\)
D \(2\,sin\,x+\dfrac {2}{x}+C\)
In some cases, it is required to split the numerator to solve the indefinite integration.
e.g. \(\displaystyle \int \dfrac{x^2+2x+1}{x+1}dx\)
\(\Rightarrow \displaystyle\int \dfrac{(x+1)(x+1)}{x+1}dx\)
\(\Rightarrow \displaystyle\int (x+1)dx\)
\(\Rightarrow \dfrac{x^2}{2} +x+C\)
A \(2\,cos\,x\)
B \(2\,sin\,x+C\)
C \(cos\,x\)
D \(sin\,x+C\)