Title: CR_Int_13
Course: Demonstrating STACK (v4.4.7)

Integration by parts

Integrate [value] with respect to [value].
Feedback: If \(u\) and \(v\) are both functions of \(x\) then \[\int u dv = uv - \int v du.\] Let \(u = [value]\) and \(dv = [value] d[value]\). Then \(du = [value] d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] \\ &\displaystyle =&\displaystyle [value] - \int [value]d[value] \\ &\displaystyle =&\displaystyle [value] + c \end{array}\]

Evaluate the following definite integral:
\[ \int^[value]_[value] [value] d[value]. \]
Feedback: If \[\int f(x) dx = F(x)\] then \[\int^a_b f(x) dx = F(a) - F(b).\] So, we need to calculate \[\int [value] d[value].\] Using integration by parts, that is if \(u\) and \(v\) are both functions of [value] then \[\int u dv = uv - \int v du,\] by defining \(u = [value]\) and \(dv = [value] d[value]\) so that \(du = [value] d[value]\) and \(v = [value]\). Then, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value]\left[[value]\right] - \int \left[[value]\right][value] d[value] \\ &\displaystyle =&\displaystyle [value] - \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] So, let \(F([value]) = [value]\) Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Evaluate the following definite integral:
\[ \int^[value]_[value] [value] d[value]. \]
Feedback: If \[\int f(x) dx = F(x)\] then \[\int^a_b f(x) dx = F(a) - F(b).\] So, we need to calculate \[\int [value] d[value].\] By linearity, \[\int [value] d[value] = [value] \int [value] d[value]\] So, using integration by parts, that is if \(u\) and \(v\) are both functions of [value] then \[\int u dv = uv - \int v du,\] by defining \(u = [value]\) and \(dv = [value] d[value]\) so that \(du = [value] d[value]\) and \(v = [value]\). Then, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value]\left[[value]\right] - \int \left[[value]\right][value] d[value] \\ &\displaystyle =&\displaystyle [value] - \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Hence, \[\int [value] d[value] = [value] = [value]\] Let \(F([value]) = [value]\) Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - [value] \end{array}\]

Integrate [value] with respect to [value].
Feedback: If \(u\) and \(v\) are both functions of [value] then \[\int u dv = uv - \int v du.\] Let \(u = [value]\) and \(dv = [value] d[value]\). Then \(du = [value] d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] \\ &\displaystyle =&\displaystyle [value] - \int [value]d[value] \end{array}\] To calculate \[\int [value]d[value]\] we will have to use parts again, by defining \(u = [value]\) and \(dv = [value]d[value]\) so that \(du = [value]d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value]d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] \\ &\displaystyle =&\displaystyle [value] - \int [value]d[value] \\ &\displaystyle =&\displaystyle [value] + c \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle [value] - \int [value]d[value] \\ &\displaystyle =&\displaystyle [value] - \left([value]\right) \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Evaluate the following definite integral:
\[ \int^[value]_[value] [value] d[value]. \]
Feedback: If \[\int f(x) dx = F(x)\] then \[\int^a_b f(x) dx = F(a) - F(b).\] Using integration by parts, \[\int u dv = uv - \int v du,\] by defining \(u = [value]\) and \(dv = [value] d[value]\) so that \(du = [value] d[value]\) and \(v = [value]\). Then, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][value] - \int [value][value] d[value] \end{array}\] To calculate \[\int [value]d[value]\] we will have to use parts again, by defining \(u = [value]\) and \(dv = [value]d[value]\) so that \(du = [value]d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value]d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][[value]] - \int [[value]][value]d[value] \\ &\displaystyle =&\displaystyle [value] - \int [value]d[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle [value] - ([value]) \\ &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Lastly, we evaluate this expression at the limits to give \[ \left[[value]\right]_[value]^[value] = [value].\]

Integrate [value] with respect to [value] using the method of integration by parts, and taking \([value]=1\cdot \ [value]\).
Feedback: If \(u\) and \(v\) are both functions of [value] then \[\int u dv = uv - \int v du.\] Let \(u = [value]\) and \(dv = d[value]\) so that \(du = [value]d[value]\) and \(v = [value]\). Then, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] \\ &\displaystyle =&\displaystyle [value] - \int [value]d[value] \\ &\displaystyle =&\displaystyle [value]+c \end{array}\]

Integrate [value] with respect to [value] using the method of integration by parts.
Feedback: To solve this problem we will have to use the method of integration by parts calc_int_methods_parts as follows: Let \(u = [value]\) and \(dv = [value]d[value]\) so that \(du = [value]d[value]\) and \(v = [value] = [value]\). Note that we can drop the modulus sign since \([value]>0\) for all the domain. Then, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] + c \\ &\displaystyle =&\displaystyle [value] - \int [value]d[value] + c \end{array}\] So, \[\begin{array}{rcl}\displaystyle 2\int [value] d[value] &\displaystyle =&\displaystyle [value] + c \\ \int [value] d[value] &\displaystyle =&\displaystyle [value] + k \end{array}\]

Integrate [value] with respect to \(t\) using the method of integration by parts.
Feedback: To solve this problem we will have to use the method of integration by parts calc_int_methods_parts two times as follows: Let \(u = [value]\) and \(dv = [value]dt\) so that \(du = [value]dt\) and \(v = [value]\). Then, \[\begin{array}{rcl}\displaystyle \int [value] dt &\displaystyle =&\displaystyle \int u dt \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][value] - \int [value] [value] dt + c \\ &\displaystyle =&\displaystyle [value] - \int [value] [value] dt + c \end{array}\] Now, define \(u = [value]\) and \(dv = [value]dt\) so that \(du = [value]dt\) and \(v = [value]\). Then, \[\begin{array}{rcl}\displaystyle \int [value][value]dt &\displaystyle =&\displaystyle \int u dt \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [value][value] - \int [value] [[value]dt] \\ &\displaystyle =&\displaystyle [value] + \int [value]dt. \end{array}\] And so \[\begin{array}{rcl}\displaystyle \int [value] dt &\displaystyle =&\displaystyle [value] - \left[[value] + \int [value] d[value]\right] + c \\ &\displaystyle =&\displaystyle [value] - [value] - \int [value] dt + c. \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle 2\int [value] dt &\displaystyle =&\displaystyle [value] + c \\ \int [value] dt &\displaystyle =&\displaystyle [value] + k. \end{array}\]