Integration by Parts, Tables, and Multivariable Calculus Study Guide

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation. Recall that the product rule is defined as $(uv)' = u'v + uv'$ or $(f \cdot g)' = f' \cdot g + f \cdot g'$. By integrating both sides of this equation, we obtain $\int (uv)' \, dx = \int u'v \, dx + \int uv' \, dx$. Solving for the integral of $uv'$, we arrive at the standard formula for Integration by Parts: $\int u \, dv = uv - \int v \, du$. For differentiable functions $u$ and $v$, this can also be expressed as $\int f(x)g'(x) \, dx = f(x)g(x) - \int f'(x)g(x) \, dx$.

Consider the example of integrating $\int x e^x \, dx$. To solve this, we define $u = x$ and $dv = e^x \, dx$. This leads to $du = 1 \, dx$ and $v = e^x$. Applying the integration by parts formula gives $x \cdot e^x - \int e^x \, dx$. Evaluating the remaining integral results in $x e^x - e^x + C$. This expression can be simplified by factoring to obtain $e^x(x - 1) + C$.

When choosing which parts of the integrand to assign to $u$ and $dv$, two general guidelines are provided. First, choose $dv$ to be the most complicated part of the integral that can be integrated easily. Second, choose $u$ so that its derivative $u'$ is simpler than $u$ itself. Specific patterns help in choosing these values for common integral forms. For integrals of the form $\int x^n e^{ax} \, dx$, choose $u = x^n$ and $dv = e^{ax} \, dx$. For integrals of the form $\int x^n \ln(x) \, dx$, choose $u = \ln(x)$ and $dv = x^n \, dx$. For integrals of the form $\int (x+a)(x+b)^n \, dx$, choose $u = x+a$ and $dv = (x+b)^n \, dx$.

In practice problem #12, the integral $\int x \ln(x) \, dx$ is evaluated by setting $u = \ln(x)$ and $dv = x \, dx$. This yields $du = \frac{1}{x} \, dx$ and $v = \frac{1}{2}x^2$. The solution is $\frac{1}{2}x^2 \ln(x) - \int \frac{1}{2}x^2 \cdot \frac{1}{x} \, dx$, which simplifies to $\frac{1}{2}x^2 \ln(x) - \frac{1}{2} \int x \, dx$, resulting in $\frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + C$. For practice problem #18, $\int (x+2)(x-5)^5 \, dx$, we set $u = x+2$ and $dv = (x-5)^5 \, dx$, leading to $du = 1 \, dx$ and $v = \frac{1}{6}(x-5)^6$. The integration yields $\frac{1}{6}(x+2)(x-5)^6 - \int \frac{1}{6}(x-5)^6 \, dx$, resulting in $\frac{1}{6}(x+2)(x-5)^6 - \frac{1}{42}(x-5)^7 + C$. Practice problem #24 integrates $\int (x+1)e^{-3x} \, dx$. Setting $u = x+1$ and $dv = e^{-3x} \, dx$ results in $du = 1 \, dx$ and $v = -\frac{1}{3}e^{-3x}$. The solution is $-\frac{1}{3}(x+1)e^{-3x} - \int -\frac{1}{3}e^{-3x} \, dx$, which simplifies to $-\frac{1}{3}(x+1)e^{-3x} - \frac{1}{9}e^{-3x} + C$.

Practice problem #26 involves $\int (x+1)^{-1/2} \ln(x+1) \, dx$. Let $u = \ln(x+1)$ and $dv = (x+1)^{-1/2} \, dx$. Thus $du = \frac{1}{x+1} \, dx$ and $v = 2(x+1)^{1/2}$. The calculation proceeds as $2(x+1)^{1/2} \ln(x+1) - \int 2(x+1)^{1/2} \cdot \frac{1}{x+1} \, dx$, which becomes $2\sqrt{x+1} \ln(x+1) - 2 \int (x+1)^{-1/2} \, dx$, resulting in $2\sqrt{x+1} \ln(x+1) - 4(x+1)^{1/2} + C$, or simplified as $2\sqrt{x+1}(\ln(x+1) - 2) + C$. Problem #28 evaluates $\int x\sqrt{x+1} \, dx$. Setting $u = x$ and $dv = (x+1)^{1/2} \, dx$ yields $du = 1 \, dx$ and $v = \frac{2}{3}(x+1)^{3/2}$. The solution is $\frac{2}{3}x(x+1)^{3/2} - \int \frac{2}{3}(x+1)^{3/2} \, dx$, resulting in $\frac{2}{3}x(x+1)^{3/2} - \frac{4}{15}(x+1)^{5/2} + C$.

Definite integrals can also be solved using these methods. Practice problem #42 evaluates $\int_{1}^{2} x \ln(x) \, dx$. Using the previous result for the indefinite integral, $\frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2$, we evaluate from $1$ to $2$. This results in $(\frac{1}{2}(2)^2 \ln(2) - \frac{1}{4}(2)^2) - (\frac{1}{2}(1)^2 \ln(1) - \frac{1}{4}(1)^2)$, which simplifies to $(2 \ln(2) - 1) - (0 - \frac{1}{4})$, resulting in a final answer of $2 \ln(2) - \frac{3}{4}$.

Integration Using Tables

Standard integral tables, often found in the back of textbooks, provide formulas for complex integrands. The steps for using an integral table involve finding a formula that matches the integrand, identifying the constants such as $a$, $b$, $c$, and $d$, and finally using the formula to complete the integration.

In problem #8, the integral is $\int \frac{15}{x^2-25} \, dx$. We identify the formula $\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln |\frac{x-a}{x+a}| + C$. Here, $a = 5$. The calculation is $15 \times \frac{1}{2(5)} \ln |\frac{x-5}{x+5}| + C$, which simplifies to $\frac{15}{10} \ln |\frac{x-5}{x+5}| + C$ or $\frac{3}{2} \ln |\frac{x-5}{x+5}| + C$. Problem #21 (listed as #11 in some sources) evaluates $\int \frac{5}{(x+1)(x+2)} \, dx$. Using the formula $\int \frac{1}{(ax+b)(cx+d)} \, dx = \frac{1}{ad-bc} (\ln|ax+b| - \ln|cx+d|) + C$, with $a=1, b=1, c=1, d=2$. The denominator constant becomes $2-1 = 1$. The result is $5 (\ln|x+2| - \ln|x+1|) + C$, which can be written as $5 \ln |x+2| - 5 \ln |x+1| + C$.

Problem #18 involves the integral $\int \frac{6}{\sqrt{4+z^2}} \, dz$. Utilizing formula #19 from the table, which fits the form $\int \frac{\sqrt{a^2+x^2}}{x} \, dx$ or $\int \frac{1}{\sqrt{a^2+x^2}} \, dx$. The transcript shows the formula application leading to $\sqrt{4+z^2} - 2 \ln |\frac{2+\sqrt{4+z^2}}{z}| + C$. Problem #30 evaluates $\int \frac{1}{9-e^{2t}} \, dt$. Let $u = e^t$ and $du = e^t \, dt$. This transforms the integral to use formula #17: $\frac{1}{2(3)} \ln |\frac{3+e^t}{3-e^t}| + C$, resulting in $\frac{1}{6} \ln |\frac{3+e^t}{3-e^t}| + C$. Problem #34 involves $\int x^2 \sqrt{x^6+1} \, dx$. Let $u = x^3$, then $du = 3x^2 \, dx$. The integral becomes $\frac{1}{3} \int \sqrt{u^2+1} \, du$, which matches formula #16.

Differential calculus and table usage are applied in real-world scenarios. In problem #67, the marginal cost for a manufacturer is given as $MC(x) = \frac{1}{\sqrt{x^2+1}}$ with fixed costs equal to $2000$, meaning $C(0) = 2000$. The cost function is the integral of the marginal cost: $C(x) = \int \frac{1}{\sqrt{x^2+1}} \, dx$. Using formula #18, we find $C(x) = \ln(x + \sqrt{x^2+1}) + K$. Setting $C(0) = 2000$, we have $2000 = \ln(0 + \sqrt{0^2+1}) + K$, which simplifies to $2000 = \ln(1) + K$. Since $\ln(1) = 0$, then $K = 2000$. The final cost function is $C(x) = \ln(x + \sqrt{x^2+1}) + 2000$.

Partial Derivatives

A partial derivative is the derivative of a function of two or more variables with respect to one variable, while treating all other variables as constants. The partial derivative of $f(x, y)$ with respect to $x$ is defined by the limit: $f_x(x, y) = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h}$. Similarly, the partial derivative with respect to $y$ is defined as: $f_y(x, y) = \lim_{h \to 0} \frac{f(x, y+h) - f(x, y)}{h}$. Subscript notation is frequently used, where $f_x(x, y)$ denotes the partial derivative with respect to $x$ and $f_y(x, y)$ denotes the partial derivative with respect to $y$.

For the function $f(x, y) = 2x^3 y^5$, the partial derivative with respect to $x$ is found by treating $y$ as a constant: $f_x(x, y) = 2(3x^2)y^5 = 6x^2 y^5$. The partial derivative with respect to $y$ is found by treating $x$ as a constant: $f_y(x, y) = 2x^3 (5y^4) = 10x^3 y^4$. In a more complex example, $f(x, y) = 2x^4 - 7x^3 y^2 - xy + 2y + 5$, the partials are calculated as follows: $f_x(x, y) = 8x^3 - 7y^2(3x^2) - y + 0 + 0$, which simplifies to $8x^3 - 21x^2 y^2 - y$. For the same function, $f_y(x, y) = 0 - 7x^3(2y) - x + 2 + 0$, which simplifies to $-14x^3 y - x + 2$.

Consider the function involving an exponential: $f(x, y) = e^{x^3 + y^2}$. To find the partial derivatives, we use the rule $\frac{d}{du}(e^u) = e^u \cdot u'$. Thus, $f_x(x, y) = e^{x^3 + y^2} \cdot 3x^2$. Evaluating at the point $(1, 2)$, we get $f_x(1, 2) = e^{1^3 + 2^2} \cdot 3(1)^2 = 3e^5$. For the partial with respect to $y$, $f_y(x, y) = e^{x^3 + y^2} \cdot 2y$. Evaluating at the same point gives $f_y(1, 2) = e^{1^3 + 2^2} \cdot 2(2) = 4e^5$.

Higher order partial derivatives involve taking subsequent derivatives. For $f(x, y) = x^3 y^4 - e^{2y}$, the first partial with respect to $x$ is $f_x = 3x^2 y^4$. The second partial $f_{xy}$ (the derivative of $f_x$ with respect to $y$) is $3x^2 (4y^3) = 12x^2 y^3$. The first partial with respect to $y$ is $f_y = 4x^3 y^3 - 2e^{2y}$. The derivative $f_{yx}$ (the derivative of $f_y$ with respect to $x$) is $4(3x^2)y^3 = 12x^2 y^3$. Note that $f_{xy} = f_{yx}$. Further derivatives, such as $f_{yxx}$, involve differentiating $f_{yx}$ with respect to $x$ again: $f_{yxx} = 12(2x)y^3 = 24xy^3$.

Optimizing Functions of Several Variables

Optimizing functions of several variables involves finding relative extrema. A function $f(x, y)$ has a relative minimum value at $(a, b)$ if the surface curves upward in all directions from that point. It has a relative maximum at $(a, b)$ if the surface curves downward in all directions. A saddle point occurs at $(a, b)$ if the function has a maximum in one direction and a minimum in another, meaning it is neither a relative maximum nor a relative minimum.

A critical point $(a, b)$ is defined as a point where both first-order partial derivatives are zero: $f_x(a, b) = 0$ and $f_y(a, b) = 0$. Relative maximum and minimum values can occur only at these critical points.

To find the critical points for the function $f(x, y) = 2x^2 + y^2 + 2xy + 4x + 2y + 5$, we first compute the partial derivatives and set them to zero. The derivative with respect to $x$ is $f_x = 4x + 2y + 4$. Setting this to zero gives $4x + 2y + 4 = 0$. The derivative with respect to $y$ is $f_y = 2y + 2x + 2$. Setting this to zero gives $2(y + x + 1) = 0$, which implies $x + y + 1 = 0$ or $y = -1 - x$. Substituting this expression for $y$ into the first equation: $4x + 2(-1 - x) + 4 = 0$. This simplifies to $4x - 2 - 2x + 4 = 0$, leading to $2x + 2 = 0$, or $x = -1$. Substituting $x = -1$ back into the expression for $y$ gives $y = -1 - (-1) = 0$. Therefore, the critical point is $(x, y) = (-1, 0)$.