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Chapter 5: Problem 8

Why can the constant of integration be omitted from the antiderivative when evaluating a definite integral?

Short Answer

Expert verified
Answer: The constant of integration can be omitted when evaluating a definite integral because it cancels out during the process of subtracting the antiderivative at the limits of integration. This general result applies to all definite integrals, making it unnecessary to include the constant of integration when evaluating them.

Step by step solution

01

Understand Antiderivatives and Constants of Integration

An antiderivative of a function is another function that represents the integral of the original function. The constant of integration is introduced because there can be an infinite number of antiderivatives for a single function. This constant represents the possible vertical shifts of the antiderivative. For example, the antiderivative of f(x) = 2x is F(x) = x^2 + C, where C is the constant of integration.
02

Understand Definite Integrals

A definite integral, written as 鈭玔a, b]f(x)dx, represents the signed area between the curve defined by the function f(x) and the x-axis, over the interval [a, b].
03

Evaluate a Definite Integral using Antiderivative

To evaluate a definite integral, we can follow the Fundamental Theorem of Calculus. It tells us to first find the antiderivative of the function, and then subtract the antiderivative evaluated at the lower limit of integration (a) from the antiderivative evaluated at the upper limit of integration (b). Mathematically: 鈭玔a, b]f(x)dx = F(b) - F(a)
04

Observe the Role of the Constant of Integration

Now, let's analyze the role of the constant of integration within the definite integral evaluation. Using the antiderivative found earlier, F(x) = x^2 + C: 鈭玔a, b]2xdx = (b^2 + C) - (a^2 + C) Notice that the constant of integration, C, appears in both terms, and when we subtract them, the C terms cancel each other out: (b^2 + C) - (a^2 + C) = b^2 - a^2
05

Conclusion

The constant of integration can be omitted from the antiderivative when evaluating a definite integral because it cancels out during the process of subtracting the antiderivative at the limits of integration. This is a general result that applies to all definite integrals, which is why we don't need to include the constant of integration when evaluating them.

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Most popular questions from this chapter

Evaluate the following integrals. $$\int_{0}^{\pi / 2} \sin ^{4} \theta d \theta$$

Find the following integrals. $$\int \frac{x}{\sqrt{x-4}} d x$$

Suppose \(f\) is continuous on \([a, b]\) with \(f^{\prime \prime}(x)>0\) on the interval. It can be shown that $$(b-a) f\left(\frac{a+b}{2}\right) \leq \int_{a}^{b} f(x) d x \leq(b-a) \frac{f(a)+f(b)}{2}$$. a. Assuming \(f\) is nonnegative on \([a, b],\) draw a figure to illustrate the geometric meaning of these inequalities. Discuss your conclusions. b. Divide these inequalities by \((b-a)\) and interpret the resulting inequalities in terms of the average value of \(f\) on \([a, b]\).

Use a change of variables to evaluate the following integrals. $$\int_{-\pi}^{0} \frac{\sin x}{2+\cos x} d x$$

A function \(f\) is symmetric about a point \((c, d)\) if whenever \((c-x, d-y)\) is on the graph, then so is \((c+x, d+y) .\) Functions that are symmetric about a point \((c, d)\) are easily integrated on an interval with midpoint \(c\). a. Show that if \(f\) is symmetric about \((c, d)\) and \(a>0,\) then \(\int_{c-a}^{c+a} f(x) d x=2 a f(c)=2 a d\). b. Graph the function \(f(x)=\sin ^{2} x\) on the interval \([0, \pi / 2]\) and show that the function is symmetric about the point \(\left(\frac{\pi}{4}, \frac{1}{2}\right)\). c. Using only the graph of \(f\) (and no integration), show that \(\int_{0}^{\pi / 2} \sin ^{2} x d x=\frac{\pi}{4} .\) (See the Guided Project Symmetry in Integrals.)
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