What is area y=sqrt(3x),y=2x-3,y=0 ?

The answer to area y=sqrt(3x),y=2x-3,y=0 is 15/4

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area y=sqrt(3x),y=2x-3,y=0

area

y = 3 x, y = 2 x - 3, y = 0

\frac{15}{4}

Decimal Notation

3.75

Solution steps

To find the intersection points solve

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

3 x = 2 x - 3 : x = 3

3 x = 0 : x = 0

2 x - 3 = 0 : x = \frac{3}{2}

= \int_{0}^{\frac{3}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{\frac{3}{2}}^{3} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{0}^{\frac{3}{2}} 3 x - 0 d x + \int_{\frac{3}{2}}^{3} 2 x - 3 - 3 x d x

\int_{0}^{\frac{3}{2}} 3 x - 0 d x = \frac{3}{2}

\int_{\frac{3}{2}}^{3} 2 x - 3 - 3 x d x = \frac{15}{4} - \frac{3}{2}

= \frac{3}{2} + \frac{15}{4} - \frac{3}{2}

Simplify

= \frac{15}{4}

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What is area y=sqrt(3x),y=2x-3,y=0 ?
The answer to area y=sqrt(3x),y=2x-3,y=0 is 15/4