What is area y=3sqrt(x),x+y=4,y=1 ?

The answer to area y=3sqrt(x),x+y=4,y=1 is 82/27

area y=3sqrt(x),x+y=4,y=1

area

y = 3 x, x + y = 4, y = 1

\frac{82}{27}

Decimal

3.03703 \dots

Solution steps

Isolate

y

for

x + y = 4 : y = 4 - x

f_{1} (x) = 3 x

f_{2} (x) = 4 - x

To find the intersection points solve

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

3 x = 4 - x : x = 1

3 x = 1 : x = \frac{1}{9}

4 - x = 1 : x = 3

= \int_{\frac{1}{9}}^{1} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{1}^{3} ∣ f_{3} (x) - f_{2} (x) ∣ d x

= \int_{\frac{1}{9}}^{1} 3 x - 1 d x + \int_{1}^{3} ∣ 1 - (4 - x) ∣ d x

\int_{\frac{1}{9}}^{1} 3 x - 1 d x = \frac{28}{27}

\int_{1}^{3} ∣ 1 - (4 - x) ∣ d x = 2

= \frac{28}{27} + 2

Simplify

= \frac{82}{27}

\int_{- 1}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + x y + y ^{2}})

y^{^{''}} - 2 y^{^{'}} - 3 y = 9

a re a 5, x^{2} + 1

\int \frac{x ^{6} - x ^{4} + x ^{2}}{x ^{3}} d x

What is area y=3sqrt(x),x+y=4,y=1 ?
The answer to area y=3sqrt(x),x+y=4,y=1 is 82/27