What is area 2y=3sqrt(x),y=3,2y+4x=7 ?

The answer to area 2y=3sqrt(x),y=3,2y+4x=7 is 41/16

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area 2y=3sqrt(x),y=3,2y+4x=7

area

2 y = 3 x, y = 3, 2 y + 4 x = 7

\frac{41}{16}

Decimal

2.5625

Solution steps

Isolate

y

for

2 y = 3 x : y = \frac{3 x}{2}

Isolate

y

for

2 y + 4 x = 7 : y = \frac{7 - 4 x}{2}

f_{1} (x) = \frac{3 x}{2}

f_{2} (x) = 3

To find the intersection points solve

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

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

\frac{3 x}{2} = \frac{7 - 4 x}{2} : x = 1

3 = \frac{7 - 4 x}{2} : x = \frac{1}{4}

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

= \int_{\frac{1}{4}}^{1} 3 - \frac{7 - 4 x}{2} d x + \int_{1}^{4} \frac{3 x}{2} - 3 d x

\int_{\frac{1}{4}}^{1} 3 - \frac{7 - 4 x}{2} d x = \frac{9}{16}

\int_{1}^{4} \frac{3 x}{2} - 3 d x = 2

= \frac{9}{16} + 2

Simplify

= \frac{41}{16}

\int_{0}^{\frac{π}{4}} 1 + tan (x) d x

\frac{d}{d x} (\frac{6 x}{x ^{2} - 9})

y^{^{'}} = (3 x - 2) (30 x + 29)

\frac{d y}{d x}, 9 x^{2} + y^{2} = 1

x \to - \infty lim (x^{2})

What is area 2y=3sqrt(x),y=3,2y+4x=7 ?
The answer to area 2y=3sqrt(x),y=3,2y+4x=7 is 41/16