Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Compute the following integrals and antiderivatives. For the definite integrals, draw a region in the plane whose area is computed by the integral. If you solve the integral by a substitution, \(u(t)=,\) then identify in writing \(u(t)\) and \(u^{\prime}(t)\) a. \(\int_{0}^{1} t^{4} d t\) b. \(\int_{0}^{1} t^{499} d t\) c. \(\int_{0}^{1} t^{1 / 2} d t\) d. \(\int_{0}^{2} e^{x} d x\) e. \(\int_{0}^{\pi} \cos z d z\) \(f . \quad \int_{2}^{6} \frac{1}{y} d y\) \(g . \quad \int_{12}^{36} \frac{1}{t} d t\) h. \(\int t^{-1 / 2} d t\) i. \(\int(\sin t+\cos t) d t\) \(j . \int \sqrt{t} d t\) k. \(\int\left(5 t^{4}+3 t^{2}+5\right) d t\) l. \(\int \frac{w^{2}+w+1}{w} d w\)

Step-by-Step Solution

Answer

a. \(\frac{1}{5}\); b. \(\frac{1}{500}\); c. \(\frac{2}{3}\); d. \(e^2 - 1\); e. 0; f. \(\ln(3)\); g. \(\ln(3)\); h. \(2t^{1/2} + C\); i. \(-\cos t + \sin t + C\); j. \(\frac{2t^{3/2}}{3} + C\); k. \(t^5 + t^3 + 5t + C\); l. \(\frac{w^2}{2} + w + \ln|w| + C\).

Step 1: Solve part a

To solve \(\int_{0}^{1} t^{4} \, dt\), we use the power rule of integration. The power rule states that the integral of \(t^n\) is \(\frac{t^{n+1}}{n+1}\). Apply the power rule:\[\int t^4 \, dt = \frac{t^{5}}{5} + C\] for constants. Evaluate from 0 to 1:\[\left. \frac{t^{5}}{5} \right|_{0}^{1} = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5}\]

Step 2: Solve part b

Similar to part a, use the power rule for integration on \(\int_{0}^{1} t^{499} \, dt\):\[\int t^{499} \, dt = \frac{t^{500}}{500} + C\]Evaluate from 0 to 1:\[\left. \frac{t^{500}}{500} \right|_{0}^{1} = \frac{1^{500}}{500} - \frac{0^{500}}{500} = \frac{1}{500}\]

Step 3: Solve part c

For \(\int_{0}^{1} t^{1/2} \, dt\), use the power rule:\[\int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} = \frac{2t^{3/2}}{3} + C\]Evaluate from 0 to 1:\[\left. \frac{2t^{3/2}}{3} \right|_{0}^{1} = \frac{2}{3} \cdot 1^{3/2} - \frac{2}{3} \cdot 0^{3/2} = \frac{2}{3}\]

Step 4: Solve part d

The integral \(\int_{0}^{2} e^{x} \, dx\) is straightforward as the integral of \(e^x\) is itself:\[ \int e^x \, dx = e^x + C\]Evaluate from 0 to 2:\[ \left. e^x \right|_{0}^{2} = e^2 - e^0 = e^2 - 1\]

Step 5: Solve part e

For \(\int_{0}^{\pi} \cos z \, dz\), the antiderivative of \(\cos z\) is \(\sin z\):\[\int \cos z \, dz = \sin z + C\]Evaluate from 0 to \(\pi\):\[ \left. \sin z \right|_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0\]

Step 6: Solve part f

To integrate \(\int_{2}^{6} \frac{1}{y} \, dy\), recognize it as \(\ln |y|\):\[\int \frac{1}{y} \, dy = \ln |y| + C\]Evaluate from 2 to 6:\[\left. \ln |y| \right|_{2}^{6} = \ln(6) - \ln(2) = \ln\left(\frac{6}{2}\right) = \ln(3)\]

Step 7: Solve part g

The integral \(\int_{12}^{36} \frac{1}{t} \, dt\) is similarly \(\ln|t|\):\[\int \frac{1}{t} \, dt = \ln |t| + C\]Evaluate from 12 to 36:\[\left. \ln |t| \right|_{12}^{36} = \ln(36) - \ln(12) = \ln\left(\frac{36}{12}\right) = \ln(3)\]

Step 8: Solve part h

For \(\int t^{-1/2} \, dt\), use the power rule:\[\int t^{-1/2} \, dt = \frac{t^{1/2}}{1/2} = 2t^{1/2} + C\]

Step 9: Solve part i

To integrate \(\int(\sin t + \cos t) \, dt\), split into two integrals:\[\int \sin t \, dt = -\cos t + C\] \[\int \cos t \, dt = \sin t + C\]Thus:\[\int(\sin t + \cos t) \, dt = -\cos t + \sin t + C\]

Step 10: Solve part j

Use the power rule for \(\int \sqrt{t} \, dt\):\[\int \sqrt{t} \, dt = \int t^{1/2} \, dt = \frac{2t^{3/2}}{3} + C\]

Step 11: Solve part k

Integrate \(\int(5t^4 + 3t^2 + 5) \, dt\) term-by-term:1. \(\int 5t^4 \, dt = \frac{5t^5}{5} = t^5 + C\)2. \(\int 3t^2 \, dt = \frac{3t^3}{3} = t^3 + C\)3. \(\int 5 \, dt = 5t + C\)Combine results:\[\int(5t^4 + 3t^2 + 5) \, dt = t^5 + t^3 + 5t + C\]

Step 12: Solve part l

First, simplify \(\frac{w^2 + w + 1}{w}\) before integrating:\( = w + 1 + \frac{1}{w}\)Now integrate each term:1. \(\int w \, dw = \frac{w^2}{2} + C\)2. \(\int 1 \, dw = w + C\)3. \(\int \frac{1}{w} \, dw = \ln|w| + C\)Combine results:\[\int \frac{w^2 + w + 1}{w} \, dw = \frac{w^2}{2} + w + \ln|w| + C\]

Step 13: Draw Regions for Definite Integrals

For definite integrals (a to g), draw the area under the curve over the specified interval on the graph. For example, part a is the area under \(t^4\) from \(t = 0\) to \(t = 1\). Similarly, do this for parts b to g with their respective functions and intervals.

Key Concepts

Definite Integrals

Definite integrals are used to calculate the area under a curve for a given interval. They essentially provide the cumulative sum of infinitesimally small values of a function over a specific range. In the exercise, you find integrals such as \(\int_{0}^{1} t^{4} \, dt\), which represents the area under the curve of \(t^4\) from \(0\) to \(1\).

• This process involves evaluating the antiderivative (integral) at the endpoints, and then taking their difference.
• For instance, finding the result of \(\left. \frac{t^{5}}{5} \right|_{0}^{1}\) gives the area under the curve.
• This results in values like \(\frac{1}{5}\), showing how the area is computed over the given interval.

Working with definite integrals helps visualize mathematical concepts and understand how function values accumulate over intervals.

Power Rule

The Power Rule is a key tool in calculus, especially for integration. It provides a straightforward method for finding antiderivatives of power functions. According to the Power Rule, the integral of \(t^n\) is \(\frac{t^{n+1}}{n+1}\). This rule plays a pivotal role in solving many of the exercises in the textbook solution.

• Applying the Power Rule involves incrementing the exponent by one and then dividing by the new exponent.
• For example, \(\int t^4 \, dt = \frac{t^5}{5} + C\), where \(C\) is the constant of integration.
• The rule simplifies the integration process, making it possible to handle more complex polynomials term-by-term.

With practice, the Power Rule becomes an automatic step in solving integration problems, allowing for quick and reliable solutions.

Antiderivatives

Antiderivatives are the reverse of derivatives. They represent the original function whose derivative matches the given function. Knowing how to find antiderivatives is essential in integral calculus as they are used to solve both definite and indefinite integrals.

• The process involves determining a function that, when differentiated, returns the original integrand.
• In the exercise, identifying antiderivatives involves expressions like \(e^x\) and \(\sin t\), where the antiderivatives are themselves or closely related functions like \(\sin x\) and \(-\cos x\), respectively.
• Every antiderivative solution includes a constant \(C\) because differentiating a constant yields zero, meaning it's undetectable in the derivative step.

Mastering antiderivatives is crucial for tackling calculus problems with confidence and precision.

Graphical Representation of Integrals

Graphical representation is a powerful tool in understanding integrals. It involves plotting a function and then shading the area under the curve over a specific interval.

• This approach provides a tangible visualization of what integrals represent, converting abstract numbers into visible areas.
• For instance, seeing the shaded region under \(t^4\) from 0 to 1 helps comprehend the calculated result as an actual area.
• It is especially useful in understanding how different functions behave and the impact of various bounds on the area.

By using graphs, students can better grasp the concept of integration and see the practical implications of the calculations they perform.