The value of int_{0}^{8} |x - 5| , dx is | vedclass

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

Question

(A) We need to evaluate the definite integral $I = \int_{0}^{8} |x - 5| \, dx$. Since the integrand $|x - 5|$ changes its definition at $x = 5$,we spl

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) We need to evaluate the definite integral $I = \int_{0}^{8} |x - 5| \, dx$. Since the integrand $|x - 5|$ changes its definition at $x = 5$,we split the integral at $x = 5$: $I = \int_{0}^{5} -(x - 5) \, dx + \int_{5}^{8} (x - 5) \, dx$. Evaluating the first part: $\int_{0}^{5} (5 - x) \, dx = [5x - \frac{x^2}{2}]_{0}^{5} = (25 - \frac{25}{2}) - 0 = \frac{25}{2} = 12.5$. Evaluating the second part: $\int_{5}^{8} (x - 5) \, dx = [\frac{x^2}{2} - 5x]_{5}^{8} = (\frac{64}{2} - 40) - (\frac{25}{2} - 25) = (32 - 40) - (12.5 - 25) = -8 - (-12.5) = 4.5$. Adding both parts: $I = 12.5 + 4.5 = 17$.

The value of $\int_{0}^{8} |x - 5| \, dx$ is

Similar Questions

Let $0 < \alpha < \beta < 1$. Then $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{dx}{1+x}$ is

$\int_0^{\pi /8} \cos^3(4\theta) \, d\theta = $

Evaluate the definite integral $\int_0^4 x[x] \, dx$,where $[x]$ denotes the greatest integer function not greater than $x$.

Choose the correct option regarding the following definite integrals:
$(i)$ $\int_0^{\pi / 2} \sin ^m(x) \cos (x) d x = \frac{1}{m+1}$
(ii) $\int_0^{\pi / 2} \sin (x) \cos ^n(x) d x = \frac{1}{n+1}$

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \,d x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module