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Pattern – 5 : Definite integration

February 27, 2021
Category: Pattern – 5 : Definite integration

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If a < c < b, then $\int_{a}^{b}f\left(x\right)dx$ = $\ \int_{a}^{c}f\left(x\right)dx+\ \int_{c}^{b}f\left(x\right)dx$

In general, a < $c_1$ < $c_2$ < $c_3$ < … < $c_n$ < b, then

$\int_{a}^{b}f\left(x\right)dx=\ \int_{a}^{c_1}f\left(x\right)dx+\ \int_{c_1}^{c_2}f\left(x\right)dx+\ \int_{c_2}^{c_3}f\left(x\right)dx+\ldots+\ \int_{c_n}^{b}f\left(x\right)\ dx$

Proof :

Let, ∫ f(x) dx = F(x) + c

LHS = $\int_{a}^{b}f\left(x\right)dx$

= $\ \left[F\left(x\right)\right]_a^b$

= F(b) – F(a) —–(1)

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

$=\ \left[F\left(x\right)\right]_a^c+\ \left[F\left(x\right)\right]_c^b$

= F(c) – F(a) + F(b) – F(c)

= F(b) – F(a) ——(2)

Question : Evaluate where, f(x) = $\begin{cases} & \text{ 1 - 4x: x }\leq 0 \\ & \text{ 1 + 4x: x }> 0 \end{cases}$

Solution : Let,

$I=\ \int_{-1}^{1}f\left(x\right)dx$

$=\ \int_{-1}^{0}f\left(x\right)dx+\ \int_{0}^{1}f\left(x\right)dx$

$=\ \int_{-1}^{0}\left(1-4x\right)dx+\ \int_{0}^{1}\left(1+4x\right)dx$

$=\ \left[x-\ \frac{4x^2}{2}\right]_{-1}^0+\ \left[x+\ \frac{4x^2}{x}\right]_0^1$

∴ I = [ ( 0 – 0 ) – (-1 – 2) ] + [ (1+2) – (0+0) ]

= 3 + 3

= 6

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Pattern – 5 : Definite integration

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