Problem 90 - Thomas Calculus

Question

A basic property of definite integrals is their invariance under translation, as expressed by the equation. $$\int_{a}^{b} f(x) d x=\int_{a-c}^{b-c} f(x+c) d x$$ The equation holds whenever \(f\) is integrable and defined for the necessary values of \(x\) . For example in the accompanying figure, show that $$\int_{-2}^{-1}(x+2)^{3} d x=\int_{0}^{1} x^{3} d x$$ because the areas of the shaded regions are congruent. (GRAPH NOT COPY) For each of the following functions, graph \(f(x)\) over \([a, b]\) and \(f(x+c)\) over \([a-c, b-c]\) to convince yourself that Equation (1) is reasonable. a. \(f(x)=x^{2}, \quad a=0, \quad b=1, \quad c=1\) b. \(f(x)=\sin x, \quad a=0, \quad b=\pi, \quad c=\pi / 2\) c. \(f(x)=\sqrt{x-4}, \quad a=4, \quad b=8, \quad c=5\)

Step-by-Step Solution

Answer

Translate and graph each function over the specified intervals to verify equivalence. Computed integrals must match for the translated function adjusted by \(c\).

Step 1: Applying the Translation Property

To demonstrate the translation property, note that \(f(x) = (x+2)^3\) is integrable over intervals satisfying the theorem's condition. We verify this for \(\int_{-2}^{-1}(x+2)^3 dx = \int_{0}^{1} x^3 dx\).

Step 2: Simplifying the Functions

For function (a), \(f(x)=x^2\), shift \(c=1\). This becomes \(f(x+1)=(x+1)^2\), effective over interval \([-1, 0]\). For function (b), \(f(x)=\sin(x)\), shift \(c=\frac{\pi}{2}\). This becomes \(f(x+\frac{\pi}{2})=\sin(x+\frac{\pi}{2})\), shifting over interval \([-\frac{\pi}{2}, 0]\). Lastly, for function (c), \(f(x)=\sqrt{x-4}\), shift \(c=5\). This results in \(f(x+5)=\sqrt{x+1}\) effective over \([-1, 3]\).

Step 3: Sketch the Graphs for Each Function

Graph each original function: for (a) over [0,1], (b) over [0,\(\pi\)] and (c) over [4,8]. Then graph each shifted function: for (a) \(f(x+1)\) over [-1,0], for (b) \(\sin(x+\frac{\pi}{2})\) over [-\(\frac{\pi}{2}\),0], and for (c), \(\sqrt{x+1}\) over [-1,3]. Make sure they visualize the congruency by highlighting the overlapping areas after translation.

Step 4: Verify the Calculations of Integration

For (a), compute the integral which confirms with transformation \(\int_{0}^{1} x^2 dx = \int_{-1}^{0} (x+1)^2 dx\). For (b), calculate \(\int_{0}^{\pi} \sin(x) dx = \int_{-\frac{\pi}{2}}^{0}\cos(x) dx\). Check (c) with \(\int_{4}^{8} \sqrt{x-4} dx = \int_{-1}^{3} \sqrt{x+1} dx\). Verify the integrations yield the same result.

Key Concepts

Definite Integrals

Definite integrals are a fundamental tool in calculus used to compute the area under a curve from one point to another along the x-axis. This involves finding the antiderivative of a function within specified limits. For instance, when we compute \( \int_a^b f(x) \, dx \), we are finding the total accumulation of the quantity represented by \( f(x) \) between the limits \( a \) and \( b \).
A definite integral can be visually interpreted as the signed area between the graph of the function and the x-axis, within the interval \([a, b]\).

• If the function is above the x-axis, the area is positive.
• If below, the area is negative.
• Different areas can cancel each other out.

The calculation of definite integrals is crucial in fields such as physics and engineering whenever a quantity needs to be accumulated over a given interval. The definite integral differs from an indefinite integral, which represents a general form of the antiderivative and includes a constant of integration.

Integration Techniques

In calculus, several integration techniques are used to evaluate both definite and indefinite integrals effectively. These techniques are crucial when dealing with complex functions that do not integrate straightforwardly through basic rules.

• Substitution: This technique involves changing variables to simplify integrals. If a function has an inner function that complicates integration, substitution can simplify the process.
• Integration by Parts: Modeled after the product rule for derivatives, it is useful for integrals of products of functions.
• Partial Fraction Decomposition: Suitable for rational functions. It breaks a complex rational function down into simpler fractions.
• Trigonometric Substitution: Helpful when dealing with square roots involving quadratic expressions, relying on trigonometric identities to simplify the integral.

Each technique requires practice to master but is essential for tackling a wide range of integral problems effectively. Using these techniques can sometimes transform difficult integrals into simpler forms that are more easily solved, as seen in the exercise through the simplification of functions for translation.

Integral Properties

Integral properties are rules that facilitate solving complex integrals by using manipulation techniques. Understanding these properties often simplifies the integration process and builds intuition about how functions behave under integration.

• Linearity: This property indicates that the integral of a sum is the sum of the integrals, and includes scalar multiplication. \[\int [a \cdot f(x) + b \cdot g(x)] \, dx = a \cdot \int f(x) \, dx + b \cdot \int g(x) \, dx\]
• Translation Property: Discussed extensively in the exercise, describes the invariance of definite integrals under horizontal shifts. Formally, \[\int_{a}^{b} f(x) \, dx = \int_{a-c}^{b-c} f(x+c) \, dx\]
• Additivity: Allows splitting the integral over adjacent intervals. \[\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx\]

These properties, when learned well, provide a solid framework for understanding how different integration problems can be approached and solved. The translation property shown in the exercise is particularly useful as it allows visualization and understanding of how integrals shift and transform under coordinate changes.