Find the area under the curve y = 9/x³ from x = 1 to x = t.

Solution:

Given, the curve y = 9/x³ from x = 1 to x = t.

We have to find the area under the curve.

Area under the curve = \(\int_{a}^{b}f(x)dx\)

Here, a = 1 and b = t.

Area under the curve = \(\int_{1}^{t}\frac{9}{x^{3}}\, dx\)

\(=9\int_{1}^{t}\frac{1}{x^{3}}\, dx\)

\(=9\int_{1}^{t}x^{-3}\, dx\)

\(=9[-\frac{1}{2}x^{-2}]_{1}^{t}\)

\(=\frac{-9}{2}[x^{-2}]_{1}^{t}\)

\(=\frac{-9}{2}[(t)^{-2}-(1)^{2}]\)

\(=\frac{-9}{2}[\frac{1}{t^{2}}-1]\)

\(=\frac{9}{2}[1-\frac{1}{t^{2}}]\)

Therefore, the area under the curve = \(=\frac{9}{2}[1-\frac{1}{t^{2}}]\).

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Find the area under the curve y = 9/x³ from x = 1 to x = t.

Summary:

The area under the curve y = 9/x³ from x = 1 to x = t is \(=\frac{9}{2}[1-\frac{1}{t^{2}}]\).