What is area 2y=5sqrt(x),y=3,2y+2x=7 ?

The answer to area 2y=5sqrt(x),y=3,2y+2x=7 is 139/600

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area 2y=5sqrt(x),y=3,2y+2x=7

area

2 y = 5 x, y = 3, 2 y + 2 x = 7

\frac{139}{600}

Decimal

0.23166 \dots

Solution steps

Isolate

y

for

2 y = 5 x : y = \frac{5 x}{2}

Isolate

y

for

2 y + 2 x = 7 : y = \frac{7 - 2 x}{2}

f_{1} (x) = \frac{5 x}{2}

f_{2} (x) = 3

To find the intersection points solve

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

\frac{5 x}{2} = 3 : x = \frac{36}{25}

\frac{5 x}{2} = \frac{7 - 2 x}{2} : x = 1

3 = \frac{7 - 2 x}{2} : x = \frac{1}{2}

= \int_{\frac{1}{2}}^{1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{1}^{\frac{36}{25}} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{\frac{1}{2}}^{1} 3 - \frac{7 - 2 x}{2} d x + \int_{1}^{\frac{36}{25}} \frac{5 x}{2} - 3 d x

\int_{\frac{1}{2}}^{1} 3 - \frac{7 - 2 x}{2} d x = \frac{1}{8}

\int_{1}^{\frac{36}{25}} \frac{5 x}{2} - 3 d x = \frac{8}{75}

= \frac{1}{8} + \frac{8}{75}

Simplify

= \frac{139}{600}

\int x (3^{x}) d x

\int (\frac{1}{- 0.4 t ^{3}}) d t

x \to 2 lim (3 (2 x + 5))

\int (x^{2} + 7 x + 3)^{4} (2 x + 7) d x

h \to 0 lim (\frac{sin ( h )}{6 h})

What is area 2y=5sqrt(x),y=3,2y+2x=7 ?
The answer to area 2y=5sqrt(x),y=3,2y+2x=7 is 139/600