Question

Find the area under the curve\(y = 1/{x^3}\)from\(x = 1\)to\(x = t\) and evaluate it for\(t = 10,100\), and 1000. Then find the total area under this curve for.

Short answer

The area under the curve is as follows:

\(\begin{aligned}{l}A(10) &= 0.495{\rm{ units}}{{\rm{ }}^2}\\A(100) &= 0.49995{\rm{ units}}{{\rm{ }}^2}\\A(1000) &= 0.4999995{\rm{ units}}{{\rm{ }}^2}\end{aligned}\)

The total area is \(\frac{1}{2}\)square units.

Step-by-step solution

Step 1:

Consider the area under the curve is determined by integrating the function over the given limit. Here, the area in actual is the area covered by the graph between the x axis and the y axis.

Consider the limit is\(a < x < b\)then, the area covered by the function\(f\left( x \right)\)is:

\(A = \int_a^b {f\left( x \right)} dx\)

Draw curve

Using graphing calculator the curve is:

Find area under the curve:

Given curve is\(y = \frac{1}{{{x^3}}}\)and the\(x\)-axis from\(x = 1\)to\(x = t\).

Integrate the function to determine the area as follows:

\(\begin{aligned}{c}A\left( t \right) &= \int_1^t {\frac{1}{{{x^3}}}} dx\\ &= \int_1^t {{x^{ - 3}}} dx\\ &= \left. {\frac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right)_1^t\\ &= \left. { - \frac{{{x^{ - 2}}}}{2}} \right)_1^t\end{aligned}\)

Solve further as:

\(\begin{aligned}{c}A\left( t \right) &= \left. { - \frac{1}{{2{x^2}}}} \right)_1^t\\ &= - \frac{1}{{2{t^2}}} - \left( { - \frac{1}{{2\left( {{1^2}} \right)}}} \right)\\ &= \frac{1}{2} - \frac{1}{{2{t^2}}}\end{aligned}\)

Evaluate

Now evaluate\(A(t) = \frac{1}{2} - \frac{1}{{2{t^2}}}\)for asked value of\(t\).

Now at \(t = 10\)

\(\begin{aligned}{c}A(10) &= \frac{1}{2} - \frac{1}{{2\left( {{{10}^2}} \right)}}\\ &= \frac{{99}}{{200}}\\ &= 0.495\end{aligned}\)

At\(t = 100\):

\(\begin{aligned}{c}A(100) &= \frac{1}{2} - \frac{1}{{2\left( {{{100}^2}} \right)}}\\ &= \frac{{9999}}{{20000}}\\ = 0.49995\end{aligned}\)

At \(t = 1000\):

\(\begin{aligned}{c}A(1000) &= \frac{1}{2} - \frac{1}{{2\left( {{{1000}^2}} \right)}}\\ &= \frac{{999999}}{{2000000}}\\ = 0.4999995\end{aligned}\)

Determine the total area

Now we are to find the total area under the curve when\(x \ge 1\), or when the upper limit\(t \to \infty \).

\(\begin{aligned}{c}\mathop {\lim }\limits_{t \to \infty } A(t) &= \int_1^\infty {\frac{1}{{{x^3}}}} dx\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{2} - \frac{1}{{2{t^2}}}} \right)\\ &= \mathop {\lim }\limits_{t \to \infty } \frac{1}{2} - \mathop {\lim }\limits_{t \to \infty } \frac{1}{{2{t^2}}}\\ &= \frac{1}{2}\end{aligned}\)