Question

Any Calc genius who wants to help me with this problem...tell me how to find the total area...i've already go the two...thanks much!

Find the area under the curve
y = 2 x^{-3}
from x = 6\ to \ x = t and evaluate it for t = 10\ , \ t = 100.
Then find the total area under this curve for x \geq 6.
(a) t = 10, (b) t = 100, (c) Total area

Answer 1

Whats x\geq6?

Answer 2

@artix_17 \geq -> \(\geq\)

Answer 3

Maybe you integrate it from 6 to infinity as an improper integral

Answer 4

@oldrin.bataku thanks

Answer 5

Let's write our area accumulating function as \(A(t)=\int_6^t2x^{-3}dx\). We can simplify this according to the fundamental theorem of calculus by writing it as a difference of the antiderivative evaluated at \(t\) and \(6\):$$A(t)=\int_6^t2x^{-3}dx=2\int_6^tx^{-3}dx=2\left(-\frac12t^{-2}+\frac126^{-2}\right)=-t^{-2}+\frac1{36}=\frac1{36}-\frac1{t^2}$$Now, let's evaluate at the given values of \(t\).
(a) \(A(10)=\frac1{36}-\frac1{100}=\frac4{225}\)
(b) \(A(100)=\frac1{36}-\frac1{10000}=\frac{2491}{90000}\)
(c) Now, we need to find the *total* area -- which should be the limit of \(A(t)\) as \(t\) approaches infinity.\[\lim_{t\to\infty}A(t)=\lim_{t\to\infty}\left(\frac1{36}-\frac1{t^2}\right)=\frac1{36}-\lim_{t\to\infty}\frac1{t^2}=\frac1{36}\]

Answer 6

@oldrin.bataku Thanks so much man!!!!!! You made my day!! :)