Problem 28 - Calculus: Early Transcendentals

Question

Evaluate the integral. \(\int_{0}^{4}(4-t) \sqrt{t} d t\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{128}{15} \).

Step 1: Identify the parts of the integral

The integral is of the form \( \int_{0}^{4}(4-t) \sqrt{t} \, d t \). We need to integrate the product of two functions: \( (4-t) \) and \( \sqrt{t} \).

Step 2: Simplify the expression

Rewrite \( \sqrt{t} \) as \( t^{1/2} \). Thus, our integral becomes: \[ \int_{0}^{4}(4-t) t^{1/2} \, d t \].

Step 3: Distribute and setup integration

Distribute \( t^{1/2} \) across \( (4-t) \) to get: \[ \int_{0}^{4} 4t^{1/2} - t^{3/2} \, d t \]. This results in two separate integrals: \[ \int_{0}^{4} 4t^{1/2} \, d t - \int_{0}^{4} t^{3/2} \, d t \].

Step 4: Integrate \( \int 4t^{1/2} \, dt \)

Integrate \( 4t^{1/2} \) with respect to \( t \):- First find the antiderivative: \( \int 4t^{1/2} \, dt = 4 \cdot \frac{t^{1/2 + 1}}{1/2 + 1} + C_1 = \frac{8}{3} t^{3/2} + C_1 \).

Step 5: Integrate \( \int t^{3/2} \, dt \)

Integrate \( t^{3/2} \) with respect to \( t \):- Find the antiderivative: \( \int t^{3/2} \, dt = \frac{t^{3/2 + 1}}{3/2 + 1} + C_2 = \frac{2}{5} t^{5/2} + C_2 \).

Step 6: Evaluate definite integral from 0 to 4

Combine and evaluate the definite integral by applying the limits:- For \( \frac{8}{3} t^{3/2} \), evaluate from 0 to 4: \[ \left[ \frac{8}{3} (4)^{3/2} \right]_0^4 = \frac{8}{3} (8) = \frac{64}{3} \]- For \( \frac{2}{5} t^{5/2} \), evaluate from 0 to 4: \[ \left[ \frac{2}{5} (4)^{5/2} \right]_0^4 = \frac{2}{5} (32) = \frac{64}{5} \]- Combine results: \( \frac{64}{3} - \frac{64}{5} = \frac{320}{15} - \frac{192}{15} = \frac{128}{15} \).

Step 7: Conclude the integration process

The value of the definite integral \( \int_{0}^{4}(4-t) \sqrt{t} \, d t \) is \( \frac{128}{15} \).

Key Concepts

Integration Techniques

Integration is a powerful tool in calculus that allows us to find areas under curves and describe accumulative processes. To compute the integral of a given function, we often need to use different techniques to make the process manageable. In the original problem, we have the integral \( \int_{0}^{4}(4-t) \sqrt{t} \, dt \), which requires a combination of methods.One of the primary techniques is simplifying the function where possible. For instance, rewriting \( \sqrt{t} \) as \( t^{1/2} \) makes the function easier to integrate. This modification reduces the problem into an integrable form where standard integration rules apply. Breaking down complex expressions into simpler components is often necessary to find their integrals.Distributing and separating integrals is another useful technique. By expressing \( (4-t) \cdot t^{1/2} \) as \( 4t^{1/2} - t^{3/2} \), we create individual simpler terms. Each can be addressed separately using basic integration formulas, which eventually simplifies the calculation process. Thus, utilizing distribution and substitution helps to handle complex integrals more effectively.

Antiderivatives

Antiderivatives, or indefinite integrals, represent the reversals of differentiation. Finding the antiderivative of a function corresponds to answering the question: "What function, when differentiated, yields the original function?"In this context, the expression \( \frac{8}{3} t^{3/2} + C_1 \) results from the integration of \( 4t^{1/2} \). Similarly, \( \frac{2}{5} t^{5/2} + C_2 \) is the antiderivative of \( t^{3/2} \). Each antiderivative involves reversing the power rule of differentiation, where we increase the power by one and divide by the new exponent.The constant \( C \) in the antiderivative is crucial in indefinite integration but does not affect definite integrals. It represents the fact that there are infinitely many functions that could have the same derivative differing only by a constant. Hence, finding antiderivatives is a fundamental aspect of resolving integrals.

Definite Integration

Definite integration computes the net area between a curve and the x-axis over a specified interval. Unlike antiderivatives, definite integrals do not result in a function but rather a number representing the total accumulation.To solve \( \int_{0}^{4} 4t^{1/2} - t^{3/2} \, dt \), we first find the antiderivatives, \( \frac{8}{3} t^{3/2} \) and \( \frac{2}{5} t^{5/2} \). The next step is to apply the limits of integration from 0 to 4 directly to these antiderivatives.By evaluating each at the upper limit and subtracting the value at the lower limit, we determine the accumulated value over \([0, 4]\). This integral evaluates to \( \frac{128}{15} \), representing the net area under the curve described by the function \((4-t) \sqrt{t}\) on this interval. Calculating the area through definite integration forms a bridge between integral calculus and real-world applications, such as physics and engineering.