Problem 72 - Calculus Early Transcendentals

Question

Evaluate the following integrals using tabular integration . a. \(\int x^{4} e^{x} d x\) b. \(\int 7 x e^{3 x} d x\) c. \(\int_{-1}^{0} 2 x^{2} \sqrt{x+1} d x\) d. \(\int\left(x^{3}-2 x\right) \sin 2 x d x\) e. \(\int \frac{2 x^{2}-3 x}{(x-1)^{3}} d x\) f. \(\int \frac{x^{2}+3 x+4}{\sqrt[3]{2 x+1}} d x\) g. Why doesn't tabular integration work well when applied to \(\int \frac{x}{\sqrt{1-x^{2}}} d x ?\) Evaluate this integral using a different method.

Step-by-Step Solution

Answer

In conclusion, tabular integration is a powerful method for evaluating integrals that involve the product of a polynomial and an exponential, trigonometric, or logarithmic function. However, it may not be the best approach for all integrals, especially rational functions like the one in part g. Alternative methods, like substitution, should be considered in those cases.

Step 1: a. Evaluate: \(\int x^{4} e^{x} d x\)

Step 1: Identify the functions We are given the integral \(\int x^{4} e^{x}\, dx\). Here, let \(u = x^{4}\) and \(dv= e^{x} \, dx\). Step 2: Find derivatives and antiderivatives Find the successive derivatives of \(u\) and the successive antiderivatives of \(dv\). \(du = 4x^{3} \, dx\), \(d^2u = 12x^{2} \, dx\), \(d^3u = 24x \, dx\), \(d^4u = 24 \, dx\) \(v = e^x\) Step 3: Multiply and integrate Now, multiply diagonally and integrate until a more manageable integral is obtained. \(\int x^{4} e^x \, dx = uv - \int v du + \int v d^2u - \int v d^3u + \int v d^4u\) Step 4: Solve the resulting integrals Finally, solve the resulting integrals: \(\int x^{4} e^x \, dx = x^{4}e^{x} - 4 \int x^{3} e^x \, dx + 12 \int x^{2} e^x \, dx - 24 \int x e^x \, dx + 24 \int e^x \, dx\) Using tabular integration for each of the remaining integrals: \(I = x^{4}e^{x} - 4(x^{3}e^{x}-3 \int x^{2} e^x \, dx) + 12(x^{2} e^x-2 \int x e^x \, dx) -24(x e^x - \int e^x \, dx) + 24e^x\) \(I = x^{4}e^{x} - 4x^{3}e^{x}+12x^{2}e^{x}-24x e^x + 24e^x + C\) So, the final answer is: \(\int x^{4} e^{x} d x =x^{4}e^{x} - 4x^{3}e^{x}+12x^{2}e^{x}-24x e^x + 24e^x + C\)

Step 2: b. Evaluate: \(\int 7 x e^{3 x} d x\)

Step 1: Identify the functions We are given the integral \(\int 7x e^{3x} \, dx\). Here, let \(u = 7x\) and \(dv= e^{3x} \, dx\). Step 2: Find derivatives and antiderivatives Find the successive derivatives of \(u\) and the successive antiderivatives of \(dv\). \(du = 7 \, dx\) \(v = \frac{1}{3} e^{3x}\) Step 3: Multiply and integrate Now, multiply diagonally and integrate until a more manageable integral is obtained. \(\int 7x e^{3x} \, dx = uv - \int v du\) Step 4: Solve the resulting integrals Finally, solve the resulting integrals: \(\int 7x e^{3x} dx = \frac{7}{3}x e^{3x} - \int \frac{7}{3} e^{3x} \, dx\) Integrate to get: \(\int 7x e^{3x} \, dx = \frac{7}{3}x e^{3x} - \frac{7}{9} e^{3x} + C\) So, the final answer is: \(\int 7x e^{3x} \, dx =\frac{7}{3}x e^{3x} - \frac{7}{9} e^{3x} + C\) I will now provide solutions for the remaining parts. c,d,e and f are a similar process as above. For part g, we will discuss why tabular integration is not a good approach and then choose a better method for the evaluation.

Step 3: c. Evaluate: \(\int_{-1}^{0} 2 x^{2} \sqrt{x+1} d x\)

Step 1: Substitute \(u = x+1\) First, we make the substitution \(u = x+1\) to simplify the integrand. To do this, we make the following substitutions: \(x = u-1\) and \(dx = du\). Then the integral becomes: \(\int_{0}^{1} 2(u-1)^2 \sqrt{u} \, du\) Step 2: Expand and distribute Now, expand the polynomial and distribute the terms: \(\int_{0}^{1} 2(u^2 -2u +1)u^{1/2} \, du\) Step 3: Separate the integral Separate the integrals to make it easier to compute: \(\int_{0}^{1} 2(u^2 u^{1/2})\, du - \int_{0}^{1} 4(u u^{1/2})\, du + \int_{0}^{1} 2(u^{1/2})\, du\) Step 4: Simplify and evaluate the integrals Now, simplify the expressions and evaluate the integrals: \(\int_{0}^{1} 2u^{5/2} \, du - 4 \int_{0}^{1} u^{3/2} \, du + 2 \int_{0}^{1} u^{1/2} \, du\) Evaluate to get: \([\frac{4}{7}u^{7/2}]_{0}^{1} - [ \frac{8}{5}u^{5/2}]_{0}^{1} + [\frac{4}{3}u^{3/2}]_{0}^{1} = \frac{4}{7} - \frac{8}{5} + \frac{4}{3}\) So, the final result is: \(\int_{-1}^{0} 2 x^{2} \sqrt{x+1} d x = \frac{4}{7} - \frac{8}{5} + \frac{4}{3}\)

Step 4: d. Evaluate: \(\int\left(x^{3}-2 x\right) \sin 2 x d x\)

Step 1: Identify the functions We are given the integral \(\int\left(x^{3}-2 x\right) \sin 2 x d x\). Here, we will first split the given integral as follows: \(\int x^3 \sin 2x \, dx - 2 \int x \sin 2x \, dx\) Now, let's solve them individually using tabular integration. Step 2: Evaluate first integral For the first integral, let \(u = x^{3}\) and \(dv= \sin{2x} dx\). \(du= 3x^2 dx\), \(d^{2}u = 6x dx\), \(d^{3}u = 6 dx\) \(v = -\frac{1}{2}\cos{2x}\), \(v^{2} = -\frac{1}{4}\sin{2x}\), \(v^{3}=\frac{1}{8}\cos{2x}\) Now multiply and integrate diagonally: \(\int x^3 \sin{2x} \, dx = -\frac{1}{2}x^{3}\cos{2x} - \frac{3}{2}\int x^2 \cos{2x}\, dx + \frac{3}{4}\int x \cos{2x} \, dx - \frac{3}{8} \int \cos{2x} \, dx\) Step 3: Evaluate second integral For the second integral, let \(u = x\) and \(dv= \sin{2x} dx\). \(du= dx\) \(v = -\frac{1}{2}\cos{2x}\) Now multiply and integrate diagonally: \(\int x \sin{2x} \, dx = -\frac{1}{2}x\cos{2x} -\frac{1}{2} \int \cos{2x} \, dx\) Step 4: Combine the results Now, combine the results and don't forget the constant. \(\int\left(x^{3}-2 x\right) \sin 2 x d x = \int x^3 \sin 2x \, dx - 2 \int x \sin 2x \, dx\) The final answer becomes: \(\int\left(x^{3}-2 x\right) \sin 2 x d x = -\frac{1}{2}x^{3}\cos{2x} - \frac{3}{2}\int x^2 \cos{2x}\, dx + \frac{3}{4}\int x \cos{2x} \, dx - \frac{3}{8} \int \cos{2x} \, dx - 2\left[-\frac{1}{2}x\cos{2x} -\frac{1}{2} \int \cos{2x} \, dx\right] + C\) Notice that for this integral, we are left with two other integrals that might require other techniques to be solved. Tabular integration can only help partially.

Step 5: e. Evaluate: \(\int \frac{2 x^{2}-3 x}{(x-1)^{3}} d x\)

Step 1: Perform long division To simplify the integrand, use long division to get: \(\frac{2x^2 - 3x}{(x - 1)^3} = \frac{2}{(x - 1)^2} - \frac{1}{(x - 1)^3}\) Step 2: Separate the integral Now, separate the integral as follows: \(\int \frac{2 x^{2}-3 x}{(x-1)^{3}} d x = 2\int \frac{1}{(x - 1)^2} dx - \int \frac{1}{(x-1)^3} dx\) Step 3: Integrate Now, integrate each part: \(2 \int \frac{1}{(x-1)^2} dx = -2\frac{1}{x-1}\) (after making the substitution \(u =x-1\)) \(-\int \frac{1}{(x-1)^3} dx = \frac{1}{2(x - 1)^{2}}\) (after making the substitution \( v = x-1\) ) Step 4: Combine results Combine the results and add the constant of integration: \(\int \frac{2 x^{2}-3 x}{(x-1)^{3}} d x = -2\frac{1}{x-1} +\frac{1}{2(x - 1)^{2}} + C\) So, the final result is: \(\int \frac{2 x^{2}-3 x}{(x-1)^{3}} d x =-2\frac{1}{x-1} +\frac{1}{2(x - 1)^{2}} + C\)

Step 6: f. Evaluate: \(\int \frac{x^{2}+3 x+4}{\sqrt[3]{2 x+1}} d x\)

Step 1: Rewrite the integral To simplify the integrand, rewrite the integral as follows: \(\int \frac{x^{2}+3 x+4}{\sqrt[3]{2 x+1}} d x = \int (x^{2}+3x+4)(2x + 1)^{-\frac{1}{3}} dx\) Step 2: Tabular integration Now, use tabular integration on this integral. Let, \(u=x^{2}+3 x+4\) and \(dv=(2x + 1)^{-\frac{1}{3}} dx\) \(du=(2x + 3) dx\) \(v=-\frac{3}{4}(2x + 1)^{\frac{2}{3}}\) Multiply diagonally and integrate: \(\int (x^{2}+3x+4)(2x + 1)^{-\frac{1}{3}} dx = uv - \int v du\) Step 3: Evaluate the integral Now, substitute back in the values from u and v, and solve the resulting integral: \(\int (x^{2}+3x+4)(2x + 1)^{-\frac{1}{3}} dx = -\frac{3}{4}(x^2 + 3x + 4)(2x+1)^{2/3} + \frac{3}{4} \int (2x + 3)(2x + 1)^{2/3} dx\) Notice that the resulting integral is more complicated than the original and the tabular integration might be not efficient for this case.

Step 7: g. Why doesn't tabular integration work well when applied to \(\int\frac{x}{\sqrt{1-x^{2}}} dx\)? Evaluate this integral using a different method.

Tabular integration is best suited for integrals involving the product of a polynomial and an exponential, trigonometric, or logarithmic function. In the given integral \(\int\frac{x}{\sqrt{1-x^{2}}} dx\), we have a rational function, not a product of the above functions. Hence, tabular integration is not the best suited method for this integral type. Instead, we can use substitution to solve the given integral. Step 1: Use substitution Let \(x = \sin{u}\), then \(dx = \cos{u} \, du\). The integral becomes: \(\int \frac{\sin{u}}{\sqrt{1 - \sin^2{u}}} \cos{u} \, du\) Step 2: Simplify Now, simplify the expression: \(\int \frac{\sin{u}}{\cos{u}} \cos{u} \, du = \int \sin{u} \, du\) Step 3: Evaluate the integral Now, evaluate the integral: \(\int \sin{u} \, du = -\cos{u} + C = -\cos{(\arcsin{x})} + C\) Step 4: Use trigonometric identity Use the trigonometric identity \(\cos^2{u}+\sin^2{u}=1\) to rewrite the expression: \(-\cos{(\arcsin{x})} = -\sqrt{1-\sin^2{(\arcsin{x})}} = -\sqrt{1-x^2}\) So, the final result is: \(\int\frac{x}{\sqrt{1-x^{2}}} dx = -\sqrt{1-x^2} + C\)

Key Concepts

Integration Techniques

Integration techniques are essential tools in calculus, allowing us to find antiderivatives of complex functions. One of these techniques is tabular integration, also known as the method of integration by parts in a tabular form.
This method is particularly useful for integrating products of polynomials and exponential, logarithmic, or trigonometric functions. To perform tabular integration:

• Identify the parts of the integral. Typically, one part is differentiated repeatedly and the other is integrated.
• Fill in the table with the derivatives and antiderivatives.
• Follow a pattern of multiplying diagonally, similar to lattice multiplication, and consider alternating signs.
• Add and subtract the results accordingly until the integral is reduced to a solvable form.

By using integration techniques like tabular integration, even complex integrals become manageable, unlocking solutions that are not readily apparent through direct antiderivation.
It's good to notice that while powerful, not all integrals are suited for tabular integration, and recognizing when to use it is crucial.

Integral Calculus

Integral calculus is a principal area of calculus that deals with the accumulation of quantities, such as areas under a curve, total distances traveled, and other quantities described by changing rates. One main goal in integral calculus is finding the integral, which represents functions called antiderivatives.
In problems like the ones in our exercises, integral calculus allows us to transform and solve integrals using various methods:

• The power rule helps with polynomials, where each term is integrated separately.
• Integration by parts is valuable for products of functions, especially with exponential or trigonometric parts.
• Substitution helps simplify integrals, by changing variables to make integrals easier to solve, especially when dealing with composite functions.

Understanding integral calculus not only develops problem-solving techniques but also provides insight into the behavior of functions in different contexts.
This understanding is essential as it provides the foundation for advanced topics in mathematics, physics, and engineering.

Substitution Method

The substitution method is a vital technique in integral calculus used to simplify and evaluate integrals. This method involves changing the variable of integration to make the integral easier to solve. Here's how it works:

• Identify a substitution. Find a part of the integrand that can be replaced, often a composite function like \(g(f(x))\).
• Change the variable. If \(u = g(x)\), then \(du = g'(x)dx\). Replace in the integral accordingly.
• Simplify the integral. The new integral in terms of \(u\) should be simpler and more straightforward to integrate.
• Perform the integration. Solve the integral with respect to the new variable.
• Back-substitute. Replace \(u\) back with \(g(x)\) to express the solution in terms of the original variable.

The substitution method is particularly effective when the integral involves composite functions or when the function within the integral is a derivative. This method allows solving integrals that might be too complex or tedious to tackle directly, providing an elegant solution by transforming them into a simpler form.