Question

Properties of integrals Use only the fact that \(\int_{0}^{4} 3 x(4-x) d x=32,\) and the definitions and properties of integrals, to evaluate the following integrals, if possible. a. \(\int_{4}^{0} 3 x(4-x) d x\) b. \(\int_{0}^{4} x(x-4) d x\) c. \(\int_{4}^{0} 6 x(4-x) d x\) d. \(\int_{0}^{8} 3 x(4-x) d x\)

Short answer

Question: Calculate the following integrals using only properties and definitions of integrals and the given value \(\int_{0}^{4} 3x(4-x) dx = 32\): a. \(\int_{4}^{0} 3x(4-x) dx\) b. \(\int_{0}^{4} x(x-4) dx\) c. \(\int_{4}^{0} 6x(4-x) dx\) d. \(\int_{0}^{8} 3x(4-x) dx\) Answer: a. \(-32\) b. \(-32\) c. \(-64\) d. \(32\)

Step-by-step solution

Reversing the limits of integration

For part (a), we will use the property of integrals that states that if we change the limits of integration, the result is multiplied by -1: \(\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx\) Hence, we can solve part (a) as follows: \(\int_{4}^{0} 3x(4-x) dx = -\int_{0}^{4} 3x(4-x) dx = -32\)

Factoring out a constant and reversing the limits of integration

For part (b), we will first factor out a -1 from the integral: \(\int_{0}^{4} x(x-4) dx = \int_{0}^{4} -1\cdot(-x^2+4x) dx\) Now, the integrand looks exactly like in the given integral except for the constant -1. We can use the property of integrals that says that constants can be taken out of the integral: \(\int_{0}^{4} -1\cdot(-x^2+4x) dx = -1\cdot\int_{0}^{4} (-x^2+4x) dx\) Now, substitute the given integral value: \(-1\cdot\int_{0}^{4} 3x(4-x) dx = -1\cdot(32)=-32\)

Multiplying the integrand by a constant

For part (c), we can use the property that says that constants inside the integrand can be moved outside the integral: \(\int_{4}^{0} 6x(4-x) dx = 2\cdot\int_{4}^{0} 3x(4-x) dx\) Now, we can use the property from Step 1 to change the limits of integration and find the value: \(2\cdot\int_{4}^{0} 3x(4-x) dx= 2\cdot(-\int_{0}^{4} 3x(4-x) dx)= -2\cdot32=-64\)

Breaking the integral into two intervals

For part (d), we will first break the integral into two intervals: \(\int_{0}^{8} 3x(4-x) dx = \int_{0}^{4} 3x(4-x) dx + \int_{4}^{8} 3x(4-x) dx\) For the first interval, we can use the given integral directly, and for the second interval, we can use the property of integrals that says that if a function is not defined for any \(x\) in the given interval, the integral will be equal to zero, since \((4-x)\) is equal to zero for all \(x\) greater than or equal to 4. Hence, the integral for part (d) will be: \(\int_{0}^{8} 3x(4-x) dx = 32 + 0 = 32\) So, the solutions of the integrals are: a. \(-32\) b. \(-32\) c. \(-64\) d. \(32\)

Key concepts

Understanding the Properties of Integrals

When evaluating integrals, certain properties can simplify the process. One important property is that reversing the limits of integration multiplies the integral by -1. This means \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \). This property is particularly useful when the limits are switched, as demonstrated in the solution to the exercise's part (a). By flipping the limits from 4 to 0, we can easily find the value of the integral by multiplying by -1.

Another key property is that constants outside the integrand can be pulled out, which simplifies many calculations. For instance, in part (b), the integration uses the fact that: \( \int_{a}^{b} c \, f(x) \, dx = c \, \int_{a}^{b} f(x) \, dx \). This allows us to factor out a "hidden" constant, simplifying the computation.

By using these properties strategically, complex-looking integrals become more manageable and easier to solve.

Grasping the Limits of Integration

The limits of integration define the range over which a function is integrated, essentially setting the "boundary" of the area you're calculating under the curve. In the current exercise, the limits of integration play a significant role in determining the final value of the integrals.

One interesting aspect is that reversing your limits changes the sign of the result. This is crucial, especially in exercises that require adjustments to integration boundaries before utilizing known integrals, as seen when solutions involve switching limits between 0 and 4.

Additionally, understanding how breaking an integral into intervals can work to your advantage is key. For example, for part (d), the integral is split over two intervals: from 0 to 4 and 4 to 8. The fundamental theorem of calculus states that the sum of integrals over subintervals is the integral over the total range. This allows for decomposition where either part becomes computationally simpler, sometimes resulting in zero, as an integral of a function not defined on some intervals.

Exploring Integration Techniques

Different integration techniques are available to solve various integral problems thoroughly. In our example, leveraging the simplicity of polynomial functions by factoring and isolating similar terms is a basic yet powerful technique.

A common technique involves constant multiplication, where solving becomes easier when taking constants outside the integral as seen in part (c). Also, the separation and recombination of integrals help manage complex problems by tackling smaller, easily solvable parts - demonstrated in part (d).

The use of expression manipulation, such as rewriting an expression in a recognizable form, can transform a daunting problem into something solvable with known values. For example, the subtraction or factoring of terms can reveal familiar structures, making computation straightforward. Practice and familiarity with these strategies greatly enhance your ability to tackle integral problems efficiently.