Question

$\int\limits^2_0 | x²+2x-3 | dx,Evaluate it.$

Answer 1

we have to evaluate , $\int\limits^2_0{|x^2+2x-3|}\,dx$

here we have considered modulus function

so first of all, we have to break modulus function.

let f(x) = |x² + 2x - 3|

for 0 < x < 1 , x² + 2x - 3 < 0

so, f(x) = -(x² + 2x - 3) , for 0 < x < 1

for 1 < x < 2, x² + 2x - 3 > 0

so, f(x) = x² + 2x - 3, for 1 < x < 2

now applying application,

$\int\limits^b_a{f(x)}\,dx=\int\limits^c_a{f(x)}\,dx+\int\limits^b_c{f(x)}\,dx$

then, integration converts into $\int\limits^1_0{-(x^2+2x-3)}\,dx-\int\limits^2_1{(x^2+2x-3)}\,dx$

= $-\left[\frac{x^3}{3}+x^2-3x\right]^1_0+\left[\frac{x^3}{3}+x^2-3x\right]^2_1$

= -[1/3 +1 - 3 - (0 + 0 - 0) ] + [(2)³/3 + (2)² - 3(2) - {(1)³/3 + (1)² - 3 }]

= -[ 4/3 - 3 ] + [8/3 + 4 - 6 - 4/3 + 3]

= -4/3 + 3 + [4/3 + 1 ]

= 4 [ Ans]

Answer 2

Answer:

Step-by-step explanation: