Title: CR_Int_14
Course: Demonstrating STACK (v4.3.7)

Integration by substitution

Integrate [value] with respect to [value] using the substitution \(u=[value]\).
Feedback: We need to calculate \[\int [value] d[value]\] by means of the substitution \(u=[value]\) calc_int_methods_substitution. So, if \(u = [value]\) then \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Integrate [value] with respect to [value] using the substitution \(u=[value]\).
Feedback: We need to calculate \[\int [value] d[value]\] by means of the substitution \(u=[value]\) [[facts:calc_int_methods_substitution]]. So, if \(u = [value]\) then \([value] = [value]\) and \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value][value] \frac{d[value]}[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Integrate [value] with respect to [value] using the substitution \(u=[value]\).
Feedback: We need to calculate \[\int [value] d[value]\] by means of the substitution \(u=[value]\) [[facts:calc_int_methods_substitution]]. So, if \(u = [value]\) then \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Integrate \[[value]\] with respect to [value] using the substitution \(z=[value]\).
Feedback: We need to solve \[\int [value] d[value]\] using the substitution \(z = [value]\) [[facts:calc_int_methods_substitution]]. So, if \(z = [value]\), then \([value] = [value]\) and \(d[value] = [value]d[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value][[value]][value]d[value] \\&\displaystyle =&\displaystyle \int [value]d[value] \\&\displaystyle =&\displaystyle [value]+c \\&\displaystyle =&\displaystyle [value] \end{array}\]

Evaluate the following definite integral: \[\int^[value]_[value] [value] d[value]\] Hint: Use the substitution \(u = [value]\)
Feedback: We need to solve \[\int [value] d[value]\] using the substitution \(u = [value]\) calc_int_methods_substitution. So, if \(u = [value]\), then \([value] = [value]\) and \(d[value] = [value]d[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [[value]] [value]d[value] \\ &\displaystyle =&\displaystyle \int [[value] d[value] \\ &\displaystyle =&\displaystyle [value]+c \\ &\displaystyle =&\displaystyle [value]+c \end{array}\] We need \[\int^[value]_[value] [value] d[value] \] so, recall that if \[\int f(x) dx = F(x)+c\] then \[\int^a_b f(x) dx = F(a) - F(b).\] 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: \[\int^[value]_[value] [value] d[value]\]
Hint: Use the substitution \(u = [value]\)
Feedback: We need to solve \[\int [value] d[value]\] using the substitution \(u = [value]\) [[facts:calc_int_methods_substitution]]. So, if \(u = [value]\), then \([value] = [value]\) and \(d[value] = [value]d[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int ([value])([value]) [value]d[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value]+c \\ &\displaystyle =&\displaystyle [value] \end{array}\] We need \[\int^[value]_[value] [value] d[value] \] so, recall that if \[\int f(x) dx = F(x)+c\] then \[\int^a_b f(x) dx = F(a) - F(b).\] 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: \[\int^[value]_[value] [value] d[value]\]
Hint: Use the substitution \(u = [value]\)
Feedback: We need to solve \[\int [value] d[value]\] using the substitution \(u = [value]\) calc_int_methods_substitution. So, if \(u = [value]\), then \(d[value] = [value]d[value]\), and so \(d[value] = \frac{d[value]}[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value]+c \\ &\displaystyle =&\displaystyle [value] \end{array}\] We need \[\int^[value]_[value] [value] d[value] \] so, recall that if \[\int f(x) dx = F(x)+c\] then \[\int^a_b f(x) dx = F(a) - F(b).\] 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: \[\int^[value]_[value] [value] d[value]\]
Hint: Use the substitution \(u = [value]\)
Feedback: We need to solve \[\int [value] d[value]\] using the substitution \(u = [value]\) calc_int_methods_substitution. So, if \(u = [value]\), then \(d[value] = [value]d[value]\), and so \[d[value] = \frac{d[value]}[value] = \frac{d[value]}[value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value]+c \\ &\displaystyle =&\displaystyle [value] \end{array}\] We need \[\int^[value]_[value] [value] d[value] \] so, recall that if \[\int f(x) dx = F(x)+c\] then \[\int^a_b f(x) dx = F(a) - F(b).\] 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: \[\int^[value]_[value] [value] d[value]\]
Hint: Use the substitution \(u = [value]\)
Feedback: We need to solve \[\int [value] d[value]\] using the substitution \(u = [value]\) calc_int_methods_substitution. So, if \(u = [value]\), then \([value] = [value]\) and \(d[value] = [value]d[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value][value][value]d[value] \\ &\displaystyle =&\displaystyle \int [value]d[value] \\ &\displaystyle =&\displaystyle [value]+c \\ &\displaystyle =&\displaystyle [value] \end{array}\] We need \[\int^[value]_[value] [value] d[value] \] so, recall that if \[\int f(x) dx = F(x)+c\] then \[\int^a_b f(x) dx = F(a) - F(b).\] So, let \[F([value]) = [value]\] Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value]-[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following integral:\[ \int [value] d[value]. \]
Feedback: We need to calculate \[\int [value] d[value]\] by means of the substitution \(u=[value]\) calc_int_methods_substitution. So, if \(u = [value]\) then \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\&\displaystyle =&\displaystyle \int [value] d[value] \\&\displaystyle =&\displaystyle [value] \\&\displaystyle =&\displaystyle [value] \end{array}\]

Find the following integral:\[ \int [value] d[value]. \]
Feedback: We need to calculate \[\int [value] d[value]\] by means of the substitution \(u=[value]\) [[facts:calc_int_methods_substitution]]. So, if \(u = [value]\) then \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\&\displaystyle =&\displaystyle \int [value] d[value] \\&\displaystyle =&\displaystyle [value] \\&\displaystyle =&\displaystyle [value] \end{array}\]

Using the method of integration by substitution, find the following integral:\[ \int [value] d[value]. \]
Feedback: We need to calculate \[\int [value] d[value]\] by means of the substitution \(u=[value]\) calc_int_methods_substitution. So, if \(u = [value]\) then \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\&\displaystyle =&\displaystyle \int [value] d[value] \\&\displaystyle =&\displaystyle [value] \int [value] d[value] \\&\displaystyle =&\displaystyle [value][value] \\&\displaystyle =&\displaystyle [value] \end{array}\]

Using the method of integration by substitution, find the following integral:\[ \int [value] d[value]. \]
Feedback: We need to calculate \[\int [value] d[value]\] by means of the substitution \(u=[value]\) [[facts:calc_int_methods_substitution]]. So, if \(u = [value]\) then \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\&\displaystyle =&\displaystyle \int [value] d[value] \\&\displaystyle =&\displaystyle [value] \int [value] d[value] \\&\displaystyle =&\displaystyle [value][value] \\&\displaystyle =&\displaystyle [value] \end{array}\]

Find the following integral using the substitution \(u=[value]\): \[ \int [value] d[value]. \]
Feedback: We need to calculate \[\int [value] d[value]\] by means of the substitution \(u=[value]\) [[facts:calc_int_methods_substitution]]. So, if \(u = [value]\) then \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value]\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{d[value]}[value] \\&\displaystyle =&\displaystyle \int [value] d[value] \\&\displaystyle =&\displaystyle [value] \\&\displaystyle =&\displaystyle [value] \end{array}\]