int_1^5 (|x-3| + |1-x|) , dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We need to evaluate the integral $I = \int_1^5 (|x-3| + |1-x|) \, dx$. Since $x \in [1, 5]$,we have $|1-x| = x-1$. Thus,$I = \int_1^5 |x-3| \, dx

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int_1^5 (|x-3| + |1-x|) \, dx$. Since $x \in [1, 5]$,we have $|1-x| = x-1$. Thus,$I = \int_1^5 |x-3| \, dx + \int_1^5 (x-1) \, dx$. For the first part,$|x-3| = 3-x$ for $x \in [1, 3]$ and $|x-3| = x-3$ for $x \in [3, 5]$. $\int_1^3 (3-x) \, dx = [3x - \frac{x^2}{2}]_1^3 = (9 - 4.5) - (3 - 0.5) = 4.5 - 2.5 = 2$. $\int_3^5 (x-3) \, dx = [\frac{x^2}{2} - 3x]_3^5 = (12.5 - 15) - (4.5 - 9) = -2.5 - (-4.5) = 2$. For the second part,$\int_1^5 (x-1) \, dx = [\frac{x^2}{2} - x]_1^5 = (12.5 - 5) - (0.5 - 1) = 7.5 - (-0.5) = 8$. Adding these results: $I = 2 + 2 + 8 = 12$.

$\int_1^5 (|x-3| + |1-x|) \, dx =$

Similar Questions

The value of $b > 3$ for which $12 \int \limits_{3}^{b} \frac{1}{(x^{2}-1)(x^{2}-4)} dx = \log _{e}(\frac{49}{40})$ is equal to

$\int_0^a {x^4 \sqrt{a^2 - x^2}} \,dx = $

$\int_0^\pi \frac{1}{1+\sin x} dx$ is equal to

$\int_{-2}^{1} [x+1] \, dx =$ (Where $[x]$ is the greatest integer function not greater than $x$)

$\int_{0}^{\frac{\pi}{4}} (\tan^n x + \tan^{n-2} x) d(x - [x])$ is : (where $[.]$ denotes the greatest integer function)

Vedclass Test Series

Exam Paper Generator

Online Exam Module