The correct evaluation of $\int_{0}^{\dfrac{\pi}{2}}|sin~(x-\dfrac{\pi}{4})\lvert.\text{d}x$ is
(A) $2+\surd2$
(B) $2-\surd2$
(C) $-2+\surd2$
(D) $0$

Hint: We can determine from the given question that the function does not have any positive values across its full range. This function's value is negative in the range of $0$ to $\dfrac{\pi}{4}$ and positive in the range of $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$. Therefore, the function is divided into two halves, one from $0$ to $\dfrac{\pi}{4}$and another from $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$.

Complete step by step solution: The given integral is: I = $\int_{0}^{\dfrac{\pi}{2}}|sin~(x-\dfrac{\pi}{4})|.\text{d}x$
Here the given function is f(x) = $\|sin~(x-\dfrac{\pi}{4})|$ and the total range over which the integral is to be done is $0$ to $\dfrac{\pi}{2}$, with lower limit $0$ and upper limit $\dfrac{\pi}{2}$. We observe that the value of this function is negative in the range of $0$ to $\dfrac{\pi}{4}$ and positive in the range of $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$.
So, the above integral can be written as I = $\int_{0}^{\dfrac{\pi}{4}}sin~(x-\dfrac{\pi}{4}).\text{d}x+\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}sin~(x-\dfrac{\pi}{4}).\text{d}x$
As integration of sin x = -cos x. So, the above integral becomes
I = $\left[-cos~(x-\dfrac{\pi}{4})\right]_{0}^{\pi/4}+\left[-cos~(x-\dfrac{\pi}{4})\right]_{\pi/4}^{\pi/2}$
We should remember that $cos 0 = 1$; cos $\dfrac{\pi}{4}$ = $\dfrac{1}{\surd2}$; cos $\dfrac{\pi}{2}$ = $0$.
Using these values to solve the question, we get
I = $(-1-(\dfrac{1}{\surd2}))+(\dfrac{1}{\surd2}-(-1))$
I = $1-\dfrac{1}{\surd2}-\dfrac{1}{\surd2}+1$
I = $2-\surd2$
Hence, the correct evaluation of $\int_{0}^{\dfrac{\pi}{2}}|sin~(x-\dfrac{\pi}{4}).\text{d}x|$ is $2-\surd2$.

Thus, Option (B) is correct.

Note: The integration is done carefully and then the upper and lower limits are substituted carefully. Integrals with no integration limit are known as indefinite integrals. Integrals with an upper and lower limit are said to be definite integrals.