What is area y=4x,y=4,y= 4/9 x^24 ?

The answer to area y=4x,y=4,y= 4/9 x^24 is 59/8

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area y=4x,y=4,y= 4/9 x^24

area

y = 4 x, y = 4, y = \frac{4}{9} x^{2} 4

\frac{59}{8}

Decimal Notation

7.375

Solution steps

To find the intersection points solve

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

4 x = 4 : x = 1

4 x = \frac{4}{9} x^{2} \cdot 4 : x = \frac{9}{4}, x = 0

4 = \frac{4}{9} x^{2} \cdot 4 : x = \frac{3}{2}, x = - \frac{3}{2}

= \int_{- \frac{3}{2}}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{\frac{3}{2}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{3}{2}}^{\frac{9}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{3}{2}}^{0} 4 - \frac{4}{9} x^{2} \cdot 4 d x + \int_{0}^{1} ∣ 4 x - 4 ∣ d x + \int_{1}^{\frac{3}{2}} ∣ 4 x - 4 ∣ d x + \int_{\frac{3}{2}}^{\frac{9}{4}} 4 x - \frac{4}{9} x^{2} \cdot 4 d x

\int_{- \frac{3}{2}}^{0} 4 - \frac{4}{9} x^{2} \cdot 4 d x = 4

\int_{0}^{1} ∣ 4 x - 4 ∣ d x = 2

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

\int_{\frac{3}{2}}^{\frac{9}{4}} 4 x - \frac{4}{9} x^{2} \cdot 4 d x = \frac{7}{8}

= 4 + 2 + \frac{1}{2} + \frac{7}{8}

Simplify

= \frac{59}{8}

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What is area y=4x,y=4,y= 4/9 x^24 ?
The answer to area y=4x,y=4,y= 4/9 x^24 is 59/8