Question

In Exercises \(13-16,\) verify that \(\int f(u) d u \neq \int f(u) d x\) $$f(u)=\sin u\( and \)u=4 x$$

Short answer

After evaluating and comparing, it's confirmed that \(\int f(u) du != \int f(u) dx\) where \(f(u) = \sin u\) and \(u = 4x\). Specifically, \(-\cos u + C != -\cos (4x)/4 + C\).

Step-by-step solution

Evaluate \(\int f(u) du\)

First, integrate the function \(f(u) = \sin u\) with respect to \(u\). The integral of \(\sin u\) with respect to \(u\) is \(-\cos u\). Therefore, \(\int f(u) du = -\cos u + C\), where \(C\) is the constant of integration.

Transform the function \(f(u)\) in terms of \(x\)

Transform the function \(f(u)\) into \(f(x)\) using the substitution \(u = 4x\). This is because the original function is given in terms of \(u\) but we want it in terms of \(x\). Therefore, \(f(x) = \sin (4x)\).

Evaluate \(\int f(u) dx\)

Next, integrate \(f(x) = \sin (4x)\) with respect to \(x\). Using the technique of substitution with \(w = 4x\), we have \(dw = 4dx\), or \(dx = dw/4\). Substituting in the integral, we get \(\int \sin w * dw/4\). The integral of sin is \(-\cos\), so \(\int \sin w * dw/4 = -\cos w/4 + C = -\cos (4x)/4 + C\).

Compare the results

Compare the results of Step 1 and Step 3. It's pretty clear to see that \(-\cos u + C != -\cos (4x)/4 + C\). Hence, it's verified that \(\int f(u) du != \int f(u) dx\).