What is area 2x+y=4,y=x,y=0 ?

The answer to area 2x+y=4,y=x,y=0 is 4/3

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area 2x+y=4,y=x,y=0

area

2 x + y = 4, y = x, y = 0

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

Isolate

y

for

2 x + y = 4 : y = 4 - 2 x

f_{1} (x) = 4 - 2 x

f_{2} (x) = x

To find the intersection points solve

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

4 - 2 x = x : x = \frac{4}{3}

4 - 2 x = 0 : x = 2

x = 0 : x = 0

= \int_{0}^{\frac{4}{3}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{4}{3}}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{4}{3}} ∣ x - 0 ∣ d x + \int_{\frac{4}{3}}^{2} ∣ 4 - 2 x - 0 ∣ d x

\int_{0}^{\frac{4}{3}} ∣ x - 0 ∣ d x = \frac{8}{9}

\int_{\frac{4}{3}}^{2} ∣ 4 - 2 x - 0 ∣ d x = \frac{4}{9}

= \frac{8}{9} + \frac{4}{9}

Simplify

= \frac{4}{3}

\int 7 e^{4 x + e^{4 x}} d x

\frac{d}{d s} (\frac{( s - 3 ) ( s - 5 )}{( s + 1 ) ( s + 2 )})

\frac{d}{d t} (3 - 2 t)

\frac{1 - x ^{2} + x ^{4}}{x ^{2} \cdot x}

x \to 0 lim (\frac{0}{0})