Question

Evaluate8 2 |x - 5|dx.​

Answer 1

Answer:

the answer ia 4x+2 if the answer is coreect pls mark me as brainliest

Answer 2

EXPLANATION.

$\sf \implies \int\limits^8_2 {| x - 5|} \, dx$

As we know that,

| x - 5 |.

There are two cases in modulus function.

⇒ (1) = | x - 5 | > 0.

⇒ x - 5 > 0.

⇒ (2) = | x - 5 | < 0.

⇒ -(x - 5) < 0.

⇒ - x + 5 < 0.

⇒ 5 - x < 0.

As we know that,

$\sf \implies \int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ where \ \ \ a < c < b.$

Using this formula in equation, we get.

$\sf \implies \int\limits^5_2 {-( x - 5)} \, dx \ + \int\limits^8_5 {(x - 5)} \, dx$

$\sf \implies \int\limits^5_2 {(5 - x)} \, dx \ + \int\limits^8_5 {(x - 5)} \, dx$

$\sf \implies \bigg[5x - \dfrac{x^{2} }{2} \bigg]_2^5 + \bigg[\dfrac{x^{2} }{2} - 5x\bigg]_5^8$

First we put the upper limit then put the lower limit.

$\sf \implies \bigg[5(5) - \dfrac{(5)^{2} }{2} \bigg] - \bigg[5(2) - \dfrac{(2)^{2} }{2} \bigg] + \bigg[\dfrac{(8)^{2} }{2} - 5(8)\bigg] - \bigg[\dfrac{(5)^{2} }{2} - 5(5) \bigg]$

$\sf \implies \bigg[25 - \dfrac{25}{2} \bigg] - \bigg[10 - \dfrac{4}{2} \bigg] + \bigg[\dfrac{64}{2} - 40 \bigg] - \bigg[\dfrac{25}{2} - 25 \bigg]$

$\sf \implies \bigg[\dfrac{50 - 25}{2} \bigg] - \bigg[ 10 - 2 \bigg] + \bigg[ 32 - 40 \bigg] - \bigg[\dfrac{25 - 50}{2} \bigg]$

$\sf \implies \bigg[\dfrac{25}{2}\bigg] - \bigg[8 \bigg] + \bigg[ -8\bigg] - \bigg[\dfrac{-25}{2} \bigg]$

$\sf \implies \dfrac{25}{2} - 8 - 8 + \dfrac{25}{2}$

$\sf \implies \dfrac{25 - 16 - 16 + 25}{2}$

$\sf \implies \dfrac{50 - 32}{2}$

$\sf \implies \dfrac{18}{2} = 9.$

$\sf \implies \int\limits^8_2 {| x - 5 | } \, dx = 9.$