# 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: $0.262592$

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: $25.060992$

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 $\int_0^1 x dx=\frac{1}{2}$ 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

$0.062592+12.460992<\int_0^1 x^6 dx+\int_1^2 x^6 dx<0.262592+25.060992$
$\mathbf{12.523584<\int_0^2 x^6 dx<25.323584}$
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

$\int_{-a}^a f(x) dx = \int_0^a f(x)dx + \int_{0}^a f(x) dx = 2\int_{0}^a f(x) dx$