# 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 ${\displaystyle f(x)=x^{6}}$ from ${\displaystyle x=0}$ to ${\displaystyle x=1}$.

${\displaystyle \left|{\begin{array}{ccccccc}i&x_{i}&x_{i}^{6}&0.2\times x_{i-1}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i-1}^{6}&0.2\times x_{i}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i}^{6}\\0&0.0&0&\,&\,&0&\,\\1&0.2&0.000064&0&0&0.0000128&0.0000128\\2&0.4&0.004096&0.0000128&0.0000128&0.0008192&0.000832\\3&0.6&0.046656&0.0008192&0.000832&0.0093312&0.0101632\\4&0.8&0.262144&0.0093312&0.0101632&0.0524288&0.062592\\5&1.0&1&0.0524288&0.062592&.2&0.262592\end{array}}\right|}$
Lower bound: ${\displaystyle 0.062592}$
Upper bound: ${\displaystyle 0.262592}$

2. Use left- and right-handed Riemann sums with 5 subdivisions to get lower and upper bounds on the area under the function ${\displaystyle f(x)=x^{6}}$ from ${\displaystyle x=1}$ to ${\displaystyle x=2}$.

${\displaystyle \left|{\begin{array}{ccccccc}i&x_{i}&x_{i}^{6}&0.2\times x_{i-1}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i-1}^{6}&0.2\times x_{i}^{6}&\sum \limits _{k=1}^{i}0.2\times x_{i}^{6}\\0&1.0&1&\,&\,&.2&\,\\1&1.2&2.985984&.2&.2&0.5971968&0.5971968\\2&1.4&7.529536&0.5971968&0.7971968&1.5059072&2.103104\\3&1.6&16.777216&1.5059072&2.303104&3.3554432&5.4585472\\4&1.8&34.012224&3.3554432&5.6585472&6.8024448&12.260992\\5&2.0&64&6.8024448&12.460992&12.8&25.060992\end{array}}\right|}$
Lower bound: ${\displaystyle 12.460992}$
Upper bound: ${\displaystyle 25.060992}$

3. Use the subtraction rule to find the area between the graphs of ${\displaystyle f(x)=x}$ and ${\displaystyle g(x)=x^{2}}$ between ${\displaystyle x=0}$ and ${\displaystyle x=1}$

From the earlier examples we know that ${\displaystyle \int _{0}^{1}xdx={\frac {1}{2}}}$ and that ${\displaystyle \int _{a}^{b}x^{2}dx={\frac {b^{3}}{3}}-{\frac {a^{3}}{3}}}$. From this we can deduce
${\displaystyle \int _{0}^{1}(x-x^{2})dx={\frac {1}{2}}-({\frac {1^{3}}{3}}-{\frac {0^{3}}{3}})={\frac {1}{2}}-{\frac {1}{3}}=\mathbf {\frac {1}{6}} }$

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 ${\displaystyle \int _{0}^{2}x^{6}dx}$.

In exercise 1 we found that

${\displaystyle 0.062592<\int _{0}^{1}x^{6}dx<0.262592}$

and in exercise 2 we found that

${\displaystyle 12.460992<\int _{1}^{2}x^{6}dx<25.060992}$

From this we can deduce that

${\displaystyle 0.062592+12.460992<\int _{0}^{1}x^{6}dx+\int _{1}^{2}x^{6}dx<0.262592+25.060992}$
${\displaystyle \mathbf {12.523584<\int _{0}^{2}x^{6}dx<25.323584} }$
5. Prove that if ${\displaystyle f}$ is a continuous even function then for any ${\displaystyle a}$,
${\displaystyle \int _{-a}^{a}f(x)dx=2\int _{0}^{a}f(x)dx.}$

From the property of linearity of the endpoints we have

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

Make the substitution ${\displaystyle u=-x;du=-dx}$. ${\displaystyle u=a}$ when ${\displaystyle x=-a}$ and ${\displaystyle u=0}$ when ${\displaystyle x=0}$. Then

${\displaystyle \int _{-a}^{0}f(x)dx=\int _{a}^{0}f(-u)(-du)=-\int _{a}^{0}f(-u)du=\int _{0}^{a}f(-u)du=\int _{0}^{a}f(u)du}$

where the last step has used the evenness of ${\displaystyle f}$. Since ${\displaystyle u}$ is just a dummy variable, we can replace it with ${\displaystyle x}$. Then

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