# Calculus/Definite integral/Solutions

1. Use left- and right-handed Riemann sums with 5 subdivisions to get lower and upper bounds on the area under the function $f(x)=x^{6}$ from $x=0$ to $x=1$ .
$\left|{\begin{array}{ccccccc}i&x_{i}&x_{i}^{6}&0.2\times x_{i-1}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i-1}^{6}&0.2\times x_{i}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i}^{6}\\0&0.0&0&\,&\,&0&\,\\1&0.2&0.000064&0&0&0.0000128&0.0000128\\2&0.4&0.004096&0.0000128&0.0000128&0.0008192&0.000832\\3&0.6&0.046656&0.0008192&0.000832&0.0093312&0.0101632\\4&0.8&0.262144&0.0093312&0.0101632&0.0524288&0.062592\\5&1.0&1&0.0524288&0.062592&.2&0.262592\end{array}}\right|$ Lower bound: $0.062592$ Upper bound: $0.262592$ $\left|{\begin{array}{ccccccc}i&x_{i}&x_{i}^{6}&0.2\times x_{i-1}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i-1}^{6}&0.2\times x_{i}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i}^{6}\\0&0.0&0&\,&\,&0&\,\\1&0.2&0.000064&0&0&0.0000128&0.0000128\\2&0.4&0.004096&0.0000128&0.0000128&0.0008192&0.000832\\3&0.6&0.046656&0.0008192&0.000832&0.0093312&0.0101632\\4&0.8&0.262144&0.0093312&0.0101632&0.0524288&0.062592\\5&1.0&1&0.0524288&0.062592&.2&0.262592\end{array}}\right|$ Lower bound: $0.062592$ Upper bound: $0.262592$ 2. Use left- and right-handed Riemann sums with 5 subdivisions to get lower and upper bounds on the area under the function $f(x)=x^{6}$ from $x=1$ to $x=2$ .
$\left|{\begin{array}{ccccccc}i&x_{i}&x_{i}^{6}&0.2\times x_{i-1}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i-1}^{6}&0.2\times x_{i}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i}^{6}\\0&1.0&1&\,&\,&.2&\,\\1&1.2&2.985984&.2&.2&0.5971968&0.5971968\\2&1.4&7.529536&0.5971968&0.7971968&1.5059072&2.103104\\3&1.6&16.777216&1.5059072&2.303104&3.3554432&5.4585472\\4&1.8&34.012224&3.3554432&5.6585472&6.8024448&12.260992\\5&2.0&64&6.8024448&12.460992&12.8&25.060992\end{array}}\right|$ Lower bound: $12.460992$ Upper bound: $25.060992$ $\left|{\begin{array}{ccccccc}i&x_{i}&x_{i}^{6}&0.2\times x_{i-1}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i-1}^{6}&0.2\times x_{i}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i}^{6}\\0&1.0&1&\,&\,&.2&\,\\1&1.2&2.985984&.2&.2&0.5971968&0.5971968\\2&1.4&7.529536&0.5971968&0.7971968&1.5059072&2.103104\\3&1.6&16.777216&1.5059072&2.303104&3.3554432&5.4585472\\4&1.8&34.012224&3.3554432&5.6585472&6.8024448&12.260992\\5&2.0&64&6.8024448&12.460992&12.8&25.060992\end{array}}\right|$ Lower bound: $12.460992$ Upper bound: $25.060992$ 3. Use the subtraction rule to find the area between the graphs of $f(x)=x$ and $g(x)=x^{2}$ between $x=0$ and $x=1$ From the earlier examples we know that $\int _{0}^{1}xdx={\frac {1}{2}}$ and that $\int _{a}^{b}x^{2}dx={\frac {b^{3}}{3}}-{\frac {a^{3}}{3}}$ . From this we can deduce
$\int _{0}^{1}(x-x^{2})dx={\frac {1}{2}}-({\frac {1^{3}}{3}}-{\frac {0^{3}}{3}})={\frac {1}{2}}-{\frac {1}{3}}=\mathbf {\frac {1}{6}}$ From the earlier examples we know that $\int _{0}^{1}xdx={\frac {1}{2}}$ and that $\int _{a}^{b}x^{2}dx={\frac {b^{3}}{3}}-{\frac {a^{3}}{3}}$ . From this we can deduce
$\int _{0}^{1}(x-x^{2})dx={\frac {1}{2}}-({\frac {1^{3}}{3}}-{\frac {0^{3}}{3}})={\frac {1}{2}}-{\frac {1}{3}}=\mathbf {\frac {1}{6}}$ 4. Use the results of exercises 1 and 2 and the property of linearity with respect to endpoints to determine upper and lower bounds on $\int _{0}^{2}x^{6}dx$ .
In exercise 1 we found that
$0.062592<\int _{0}^{1}x^{6}dx<0.262592$ and in exercise 2 we found that

$12.460992<\int _{1}^{2}x^{6}dx<25.060992$ From this we can deduce that

$0.062592+12.460992<\int _{0}^{1}x^{6}dx+\int _{1}^{2}x^{6}dx<0.262592+25.060992$ $\mathbf {12.523584<\int _{0}^{2}x^{6}dx<25.323584}$ In exercise 1 we found that
$0.062592<\int _{0}^{1}x^{6}dx<0.262592$ and in exercise 2 we found that

$12.460992<\int _{1}^{2}x^{6}dx<25.060992$ From this we can deduce that

$0.062592+12.460992<\int _{0}^{1}x^{6}dx+\int _{1}^{2}x^{6}dx<0.262592+25.060992$ $\mathbf {12.523584<\int _{0}^{2}x^{6}dx<25.323584}$ 5. Prove that if $f$ is a continuous even function then for any $a$ ,
$\int _{-a}^{a}f(x)dx=2\int _{0}^{a}f(x)dx.$ From the property of linearity of the endpoints we have
$\int _{-a}^{a}f(x)dx=\int _{-a}^{0}f(x)dx+\int _{0}^{a}f(x)dx$ Make the substitution $u=-x;du=-dx$ . $u=a$ when $x=-a$ and $u=0$ when $x=0$ . Then

$\int _{-a}^{0}f(x)dx=\int _{a}^{0}f(-u)(-du)=-\int _{a}^{0}f(-u)du=\int _{0}^{a}f(-u)du=\int _{0}^{a}f(u)du$ where the last step has used the evenness of $f$ . Since $u$ is just a dummy variable, we can replace it with $x$ . Then

$\int _{-a}^{a}f(x)dx=\int _{0}^{a}f(x)dx+\int _{0}^{a}f(x)dx=2\int _{0}^{a}f(x)dx$ From the property of linearity of the endpoints we have
$\int _{-a}^{a}f(x)dx=\int _{-a}^{0}f(x)dx+\int _{0}^{a}f(x)dx$ Make the substitution $u=-x;du=-dx$ . $u=a$ when $x=-a$ and $u=0$ when $x=0$ . Then

$\int _{-a}^{0}f(x)dx=\int _{a}^{0}f(-u)(-du)=-\int _{a}^{0}f(-u)du=\int _{0}^{a}f(-u)du=\int _{0}^{a}f(u)du$ where the last step has used the evenness of $f$ . Since $u$ is just a dummy variable, we can replace it with $x$ . Then

$\int _{-a}^{a}f(x)dx=\int _{0}^{a}f(x)dx+\int _{0}^{a}f(x)dx=2\int _{0}^{a}f(x)dx$ 