What is area 2x,5, 2/(x^2) ?

The answer to area 2x,5, 2/(x^2) is diverges-2sqrt(10)+37/4

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area 2x,5, 2/(x^2)

area

2 x, 5, \frac{2}{x ^{2}}

diverges - 210 + \frac{37}{4}

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

2 x = \frac{2}{x ^{2}} : x = 1

2 x = 5 : x = \frac{5}{2}

\frac{2}{x ^{2}} = 5 : x = \frac{2}{5}, x = - \frac{2}{5}

= \int_{- \frac{2}{5}}^{\frac{2}{5}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{2}{5}}^{1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{1}^{\frac{5}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{2}{5}}^{\frac{2}{5}} \frac{2}{x ^{2}} - 5 d x + \int_{\frac{2}{5}}^{1} \frac{2}{x ^{2}} - 5 d x + \int_{1}^{\frac{5}{2}} ∣ 2 x - 5 ∣ d x

\int_{- \frac{2}{5}}^{\frac{2}{5}} \frac{2}{x ^{2}} - 5 d x =

diverges

\int_{\frac{2}{5}}^{1} \frac{2}{x ^{2}} - 5 d x = 7 - 210

\int_{1}^{\frac{5}{2}} ∣ 2 x - 5 ∣ d x = \frac{9}{4}

= diverges + 7 - 210 + \frac{9}{4}

Simplify

= diverges - 210 + \frac{37}{4}

x \to \infty lim (2 x - 4 x^{2} - 2 x)

\int \frac{1}{2 x + 2} d x

5^{t}

\int 2 z ln (y) d z

\int \frac{15}{x} d x

What is area 2x,5, 2/(x^2) ?
The answer to area 2x,5, 2/(x^2) is diverges-2sqrt(10)+37/4