What is area y=5sqrt(x),y=4,2y+x=6 ?

The answer to area y=5sqrt(x),y=4,2y+x=6 is -250sqrt(31)+104464/75+(20sqrt(2)(28-5sqrt(31))^{3/2})/3

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area y=5sqrt(x),y=4,2y+x=6

area

y = 5 x, y = 4, 2 y + x = 6

- 25031 + \frac{104464}{75} + \frac{20 2 ( 28 - 5 31 ) ^{\frac{3}{2}}}{3}

Decimal

1.52231 \dots

Solution steps

Isolate

y

for

2 y + x = 6 : y = \frac{6 - x}{2}

f_{1} (x) = 5 x

f_{2} (x) = 4

To find the intersection points solve

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

5 x = 4 : x = \frac{16}{25}

5 x = \frac{6 - x}{2} : x = 2 (28 - 531)

4 = \frac{6 - x}{2} : x = - 2

= \int_{- 2}^{2 (28 - 531)} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{2 (28 - 531)}^{\frac{16}{25}} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 2}^{2 (28 - 531)} 4 - \frac{6 - x}{2} d x + \int_{2 (28 - 531)}^{\frac{16}{25}} 5 x - 4 d x

\int_{- 2}^{2 (28 - 531)} 4 - \frac{6 - x}{2} d x = 1616 - 29031

\int_{2 (28 - 531)}^{\frac{16}{25}} 5 x - 4 d x = 4031 - \frac{16736}{75} + \frac{20 2 ( 28 - 5 31 ) ^{\frac{3}{2}}}{3}

= 1616 - 29031 + 4031 - \frac{16736}{75} + \frac{20 2 ( 28 - 5 31 ) ^{\frac{3}{2}}}{3}

Simplify

= - 25031 + \frac{104464}{75} + \frac{20 2 ( 28 - 5 31 ) ^{\frac{3}{2}}}{3}

\int_{0}^{1} x 1 + x^{2} d x

x^{6} (1 - \frac{1}{x + 5})

\int_{1}^{8} \frac{7 ln ( x )}{x ^{2}} d x

x \to 2 lim (\frac{1}{x} - \frac{1}{x ^{2}})

\frac{π}{6}

What is area y=5sqrt(x),y=4,2y+x=6 ?
The answer to area y=5sqrt(x),y=4,2y+x=6 is -250sqrt(31)+104464/75+(20sqrt(2)(28-5sqrt(31))^{3/2})/3