91Ó°ÊÓ

We are given the integral: $$\int_{0}^{3 \pi / 8} \sin \left(2 x-\frac{\pi}{4}\right) d x$$ The function to integrate is \(\sin \left(2 x-\frac{\pi}{4}\right)\), and the limits of integration are \(0\) and \(3\pi/8\).

Using basic rules of integration, we can find the antiderivative of the given function. The antiderivative of \(\sin(ax+b)\), with respect to \(x\), is \(-\frac{1}{a}\cos(ax+b)\). So the antiderivative of \(\sin \left(2 x-\frac{\pi}{4}\right)\) is: $$F(x) = -\frac{1}{2} \cos \left(2 x - \frac{\pi}{4}\right) + C$$ Note that we don't need to include the constant of integration in this case, since we will be evaluating the definite integral.

Now that we have the antiderivative, we can apply the fundamental theorem of calculus to evaluate the definite integral. Using the limits \(0\) and \(3\pi/8\), we have: $$\int_{0}^{3 \pi / 8} \sin \left(2 x-\frac{\pi}{4}\right) d x = F\left(\frac{3 \pi}{8}\right) - F(0)$$ Substitute the values into the antiderivative expression: $$\left[-\frac{1}{2} \cos \left(2 \cdot \frac{3 \pi}{8} - \frac{\pi}{4}\right)\right] - \left[-\frac{1}{2} \cos \left(2 \cdot 0 - \frac{\pi}{4}\right)\right]$$ Now, simplify the expression: $$\left[-\frac{1}{2} \cos \left(\frac{3 \pi}{4}\right)\right] - \left[-\frac{1}{2} \cos \left(-\frac{\pi}{4}\right)\right]$$ Finally, calculate the value of the cosine functions: $$\left[-\frac{1}{2} \cdot 0 \right] - \left[-\frac{1}{2} \cdot \frac{\sqrt{2}}{2}\right]$$ So the value of the integral is: $$\boxed{\frac{\sqrt{2}}{4}}$$