Question

Given \(\int_{-1}^{1} f(x) d x=0\) and \(\int_{0}^{1} f(x) d x=5,\) evaluate (a) \(\int_{-1}^{0} f(x) d x\). (b) \(\int_{0}^{1} f(x) d x-\int_{-1}^{0} f(x) d x\) (c) \(\int_{-1}^{1} 3 f(x) d x\). (d) \(\int_{0}^{1} 3 f(x) d x\).

Short answer

\((a) -5, (b) 10, (c) 0, (d) 15.\)

Step-by-step solution

Split the integral over the total interval

The integral from -1 to 1 could be represented as the sum of the integral from -1 to 0 and the integral from 0 to 1, i.e., \(\int_{-1}^{1} f(x) d x = \int_{-1}^{0} f(x) d x + \int_{0}^{1} f(x) d x\).

Solve the first part of the problem (a)

Substituting the given values into the equation from step 1, we get \(0 = \int_{-1}^{0} f(x) d x + 5\), then by algebraically rearranging, we find \(\int_{-1}^{0} f(x) d x = -5\).

Solve the second part of the problem (b)

\(\int_{0}^{1} f(x) d x - \int_{-1}^{0} f(x) d x = 5 - (-5) = 10\).

Solve the third part of the problem (c)

Use the scalar multiplication property of integrals. \(\int_{-1}^{1} 3 f(x) d x = 3 \int_{-1}^{1} f(x) d x = 3*0 = 0\).

Solve the final part of the problem (d)

As above, using the scalar multiplication property of integrals, \(\int_{0}^{1} 3 f(x) d x = 3 \int_{0}^{1} f(x) d x = 3*5 = 15\).

Key concepts

Definite Integrals

The concept of definite integrals is an essential building block in understanding calculus. It represents the accumulation of quantities, like area under a curve, between two points on a graph. In mathematical terms, a definite integral is written as \(\int_{a}^{b} f(x) dx\), where \(f(x)\) is the function to be integrated, and \(a\) and \(b\) are the lower and upper limits of the integration, respectively.

For example, if you have an integral like \(\int_{-1}^{1} f(x) dx = 0\), it means that the net area calculated by the function \(f(x)\) from \(x = -1\) to \(x = 1\) is zero. This could result if the function is symmetric about the y-axis, or if the areas above and below the x-axis cancel each other out. Understanding this principle is crucial to evaluate integrals correctly and to apply them to real-world problems.

Properties of Integrals

Additivity of Integrals

The example we have involves splitting an integral over an interval into two integrals. This property of additivity states that \(\int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx\), given that \(a < b < c\). This property can simplify the calculation when we know values for parts of the interval.

Scalar Multiplication

Another useful property showcased is scalar multiplication, which allows us to take a constant multiplier outside of the integral sign, that is, \(\int k\cdot f(x) dx = k \int f(x) dx\). It's instrumental when dealing with integrals within a real-world context, where constants may represent scaling factors like density or speed.

Integral Evaluation

Evaluating an integral involves finding the value of the integral, often with respect to certain limits. In our example, the evaluation of different integral expressions requires both the properties of integrals and algebraic manipulation.

The process involves substituting known values and rearranging the equations to find the values of unknown integrals. For instance, to solve for \(\int_{-1}^{0} f(x) dx\), knowing that \(\int_{-1}^{1} f(x) dx = 0\) and \(\int_{0}^{1} f(x) dx = 5\), we use these values as pieces of a puzzle to uncover the integral over the unknown interval. The algebra plays a crucial role in arriving at the correct answer efficiently.