Calculus/Definite integral/Solutions

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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