What is area 2y=3sqrt(x),y=4,2y+3x=6 ?

The answer to area 2y=3sqrt(x),y=4,2y+3x=6 is 925/108

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area 2y=3sqrt(x),y=4,2y+3x=6

area

2 y = 3 x, y = 4, 2 y + 3 x = 6

\frac{925}{108}

Decimal Notation

8.56481 \dots

Solution steps

Isolate

y

for

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

Isolate

y

for

2 y + 3 x = 6 : y = \frac{6 - 3 x}{2}

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

f_{2} (x) = 4

To find the intersection points solve

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

\frac{3 x}{2} = 4 : x = \frac{64}{9}

\frac{3 x}{2} = \frac{6 - 3 x}{2} : x = 1

4 = \frac{6 - 3 x}{2} : x = - \frac{2}{3}

= \int_{- \frac{2}{3}}^{1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{1}^{\frac{64}{9}} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- \frac{2}{3}}^{1} 4 - \frac{6 - 3 x}{2} d x + \int_{1}^{\frac{64}{9}} \frac{3 x}{2} - 4 d x

\int_{- \frac{2}{3}}^{1} 4 - \frac{6 - 3 x}{2} d x = \frac{25}{12}

\int_{1}^{\frac{64}{9}} \frac{3 x}{2} - 4 d x = \frac{175}{27}

= \frac{25}{12} + \frac{175}{27}

Simplify

= \frac{925}{108}

\int \frac{( x - 1 )}{( ( x - 2 ) ( x ^{2} - 2 x + 2 ) ^{2} )} d x

x \to 0 + lim ((3 x)^{s i n (x)})

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

\int 5 sin^{3} (x) cos^{7} (x) d x

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

What is area 2y=3sqrt(x),y=4,2y+3x=6 ?
The answer to area 2y=3sqrt(x),y=4,2y+3x=6 is 925/108