Problem 16 - Calculus: Graphical, Numerical, Algebraic

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$$

Step-by-Step Solution

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 1: 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.

Step 2: 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, \(f(x) = \sin (4x)\).

Step 3: 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 \(\int \sin w * dw/4 = -\cos w/4 + C = -\cos (4x)/4 + C\).

Step 4: 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 \(\int f(u) du != \int f(u) dx\).

Key Concepts

Definite Integrals

Definite integrals are used to find the area under a curve between specified limits, say from point \(a\) to point \(b\). This is an essential concept in integral calculus because instead of just finding a family of functions as with indefinite integrals, it results in a real number representing the accumulation of quantities. In the case of definite integrals, there are no constants of integration like \(C\), which is often present in indefinite integrals.
When you calculate a definite integral, you must evaluate two values: you compute the integral of the function and then evaluate it at the upper and lower limits. The result of these evaluations and their difference gives the accumulated value that the definite integral represents. This method is widely employed in physics and engineering for solving real-world problems involving areas and volumes.

Substitution Method

The substitution method is a powerful technique for evaluating integrals that might not be easily solvable in their current form. It's especially useful when dealing with composite functions, where one function is nested inside another.
The main idea with substitution is to simplify the integral by changing variables. You essentially "swap" the difficult part out for something simpler. For example, in our problem, we substituted \(u = 4x\) to transform the function \(f(x) = \sin(4x)\) into something more manageable. This transformation allowed us to have a clear path to integrate.
Once you apply the substitution, you also need to change the differential. As in our example, if \(w = 4x\), then \(dw = 4dx\) and hence \(dx = dw/4\). After integrating in terms of the new variable, remember to substitute back the original variable to complete the process. This method makes complex integrals more approachable and extends your ability to solve a wide range of problems.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include trigonometrical expressions like \(\sin(x)\), \(\cos(x)\), and \(\tan(x)\). They can become complicated due to the periodic nature of trigonometric functions.
Integrating these functions often requires familiarity with specific integral formulas, similar to the integral of \(\sin(u)\) which is \(-\cos(u) + C\). Recognizing these basic integrals quickly can save a lot of time. In cases like \(\int \sin(4x) \, dx\), it might be necessary to employ substitution methods as seen in our problem, to simplify the product of more complex trigonometric expressions.
Practicing trigonometric integrals hones your skills not only for identifying complex functions quickly, but also in transforming them effectively for integration using identities and substitutions. This makes them a crucial part of integral calculus.