Question

Find 2 to -2 integration (|x+1|+|x|)dx is the value integral

Answer 1

we have to find $\int\limits^2_{-2}{(|x+1|+|x|)}\,dx$

solution : break the modulus function and find intervals.

|x + 1| = 0 ⇒x = -1

|x| = 0 ⇒x = 0

so intervals are -2 to -1 , -1 to 0 and 0 to 2.

case 1 : -2 < x < -1

|x + 1| + |x| = -x - 1 - x = -2x - 1

case 2 : - 1 < x < 0

|x + 1| + |x| = x + 1 - x = 1

case 3 : 0 < x < 2

|x + 1| + |x| = x + 1 + x = 2x + 1

now $\int\limits^2_{-2}{(|x+1|+|x|)}\,dx$

= $\int\limits^2_0{(2x+1)}\,dx+\int\limits^0_{-1}{(1)}\,dx+\int\limits^{-1}_{-2}{(-2x-1)}\,dx$

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

= (2² + 2) + (1) -[(-1)² + (-1) - (-2)² - (-2)]

= 6 + 1 - [1 - 1 - 4 + 2 ]

= 6 + 1 + 2

= 9

Therefore the value of $\int\limits^2_{-2}{(|x+1|+|x|)}\,dx$ is 9.

Answer 2

Step-by-step explanation:

we have to find \int\limits^2_{-2}{(|x+1|+|x|)}\,dx−2∫2(∣x+1∣+∣x∣)dx

solution : break the modulus function and find intervals.

|x + 1| = 0 ⇒x = -1

|x| = 0 ⇒x = 0

so intervals are -2 to -1 , -1 to 0 and 0 to 2.

case 1 : -2 < x < -1

|x + 1| + |x| = -x - 1 - x = -2x - 1

case 2 : - 1 < x < 0

|x + 1| + |x| = x + 1 - x = 1

case 3 : 0 < x < 2

|x + 1| + |x| = x + 1 + x = 2x + 1

now \int\limits^2_{-2}{(|x+1|+|x|)}\,dx−2∫2(∣x+1∣+∣x∣)dx

= \int\limits^2_0{(2x+1)}\,dx+\int\limits^0_{-1}{(1)}\,dx+\int\limits^{-1}_{-2}{(-2x-1)}\,dx0∫2(2x+1)dx+−1∫0(1)dx+−2∫−1(−2x−1)dx

= \left[x^2+x\right]^2_0+\left[x\right]^{-1}_0-\left[x^2+x\right]^{-1}_{-2}[x2+x]02+[x]0−1−[x2+x]−2−1

= (2² + 2) + (1) -[(-1)² + (-1) - (-2)² - (-2)]

= 6 + 1 - [1 - 1 - 4 + 2 ]

= 6 + 1 + 2

= 9

Therefore the value of \int\limits^2_{-2}{(|x+1|+|x|)}\,dx−2∫2(∣x+1∣+∣x∣)dx is 9.