What is area x^2,6x-9,y=0 ?

The answer to area x^2,6x-9,y=0 is 9/4

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area x^2,6x-9,y=0

area

x^{2}, 6 x - 9, y = 0

\frac{9}{4}

Decimal Notation

2.25

Solution steps

To find the intersection points solve

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

0 = x^{2} : x = 0

0 = 6 x - 9 : x = \frac{3}{2}

x^{2} = 6 x - 9 : x = 3

= \int_{0}^{\frac{3}{2}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{3}{2}}^{3} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{3}{2}} 0 - x^{2} d x + \int_{\frac{3}{2}}^{3} x^{2} - (6 x - 9) d x

\int_{0}^{\frac{3}{2}} 0 - x^{2} d x = \frac{9}{8}

\int_{\frac{3}{2}}^{3} x^{2} - (6 x - 9) d x = \frac{9}{8}

= \frac{9}{8} + \frac{9}{8}

Simplify

= \frac{9}{4}

y^{^{'}} - \frac{2}{x} y = \frac{y ^{3}}{x ^{3}}, y (1) = 1

- \frac{2}{5} x^{2}, at x = - 1

- \frac{cos ( \frac{1}{x} )}{x ^{2}}

y = e^{x}, (0, 1)

\frac{6}{3 x + 1}