What is area y=x,y=1-x,y=-x/2 ?

The answer to area y=x,y=1-x,y=-x/2 is 3/4

area y=x,y=1-x,y=-x/2

area

y = x, y = 1 - x, y = - \frac{x}{2}

\frac{3}{4}

Decimal

0.75

Solution steps

To find the intersection points solve

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

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

x = - \frac{x}{2} : x = 0

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

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

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

\int_{0}^{\frac{1}{2}} x - (- \frac{x}{2}) d x = \frac{3}{16}

\int_{\frac{1}{2}}^{2} - \frac{x}{2} - (1 - x) d x = \frac{9}{16}

= \frac{3}{16} + \frac{9}{16}

Simplify

= \frac{3}{4}

\int \frac{1}{3 x lo g _{7} ( x )} d x

\frac{\partial}{\partial y} (e^{x} + y)

\frac{\partial}{\partial x} (x^{2} - 5 x y + 3 y^{2} + x - 4 y)

\int (x x^{4}) d x

y^{^{'''}} - 5 y^{^{''}} + 8 y^{^{'}} + 14 y = 0