Question

In Exercises \(7-12,\) evaluate the integral. $$\int_{0}^{3}(-160) d t$$

Short answer

The value of the integral is -480.

Step-by-step solution

Identify the constant term

In this integral, the constant term is -160.

Identify the limits of integration

The limits of integration are from 0 to 3.

Apply the integral rule for constants

The integral of a constant \(c\) from \(a\) to \(b\) is \(c*(b - a)\). In this case, \(c\) is -160, \(a\) is 0 and \(b\) is 3. So, the integral is -160 * (3 - 0) = -160 * 3.

Calculate the result

The final result is -480.

Key concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that deals with the accumulation of quantities and the area under and between curves. It is divided into two main types: indefinite integrals, which represent a family of functions, and definite integrals, which evaluate to a real number representing the accumulated value between two points on a graph. The process involves finding the antiderivative of a function and then applying the limits of integration.

More specifically, when we talk about evaluating definite integrals, we're calculating the net area under a curve from one point to another. In the context of the exercise provided, we're looking at a constant function, which simplifies the integration process. For constant functions, the integral essentially becomes a multiplication problem, taking the constant and multiplying it by the difference in the limits of integration.

Constant Function Integration

Integration of a constant function is one of the simplest cases in integral calculus. When the integrand, which is the function inside the integral sign, is a constant, the integral represents the area of a rectangle where the height is the constant value and the width is the interval between the limits of integration.

In mathematical terms, if you have a constant function represented as \(f(t) = c\), the integral from \(a\) to \(b\) is given as \(\int_{a}^{b} c \, dt = c \times (b - a)\). This formula reflects that you're multiplying the height (the constant \(c\)) by the width (the difference between \(b\) and \(a\)). In our exercise, the constant was -160, and applying this concept leads to a quick evaluation, which is particularly useful when dealing with constants over larger intervals.

Limits of Integration

The limits of integration are crucial in defining the interval over which we're integrating. They tell us where to start and end our calculation. In a definite integral, these are typically represented by the numbers at the top and bottom of the integral sign. The lower limit, or the bottom number, is where we start, and the upper limit, or the top number, is where we stop.

Understanding the impact of these limits can greatly influence the outcome of the integral. Reversing the limits, for instance, would change the sign of the result. In our exercise, the limits of integration were from 0 to 3; thus, we multiply our constant by 3. Had the limits been reversed, from 3 to 0, the result would have been a positive 480 instead of -480, demonstrating the importance of carefully noting the order of integration limits.