∫04 |x−1|dx is equal to - Infinity Learn by Sri Chaitanya

A
$\frac{5}{2}$
B
$\frac{3}{2}$
C
$\frac{1}{2}$
D
5

Solution:

Let $\int_{0}^{4} | x - 1 | dx$

It can be seen that, $(x - 1) \leq 0$ when $0 \leq x \leq 1$ and

$(x - 1) \geq 0 when 1 \leq x \leq 4$

$\begin{array}{l} ∴ I = \int_{0}^{1} | x - 1 | dx + \int_{1}^{4} | x - 1 | dx \\ = \int_{0}^{1} (1 - x) dx + \int_{a}^{b} f (x) dx = \int_{a}^{c} f (x) dx + \int_{c}^{b} f (x) dx] \\ = {[x - \frac{x^{2}}{2}]}_{0}^{1} + {[\frac{x^{2}}{2} - x]}_{1}^{4} \\ = (1 - \frac{1}{2}) - 0 + (\frac{4^{2}}{2} - 4) - (\frac{1}{2} - 1) = \frac{1}{2} + 4 + \frac{1}{2} = 5 \end{array}$

$\int_{0}^{4} | x - 1 | dx$ is equal to

Solution:

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