What is area 5x, 5/3 x,14-x^2 ?

The answer to area 5x, 5/3 x,14-x^2 is 5405/81

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area 5x, 5/3 x,14-x^2

area

5 x, \frac{5}{3} x, 14 - x^{2}

\frac{5405}{81}

Decimal Notation

66.72839 \dots

Solution steps

To find the intersection points solve

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

5 x = \frac{5}{3} x : x = 0

5 x = 14 - x^{2} : x = - 7, x = 2

\frac{5}{3} x = 14 - x^{2} : x = - \frac{14}{3}, x = 3

= \int_{- 7}^{- \frac{14}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{14}{3}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{2} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{2}^{3} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 7}^{- \frac{14}{3}} 5 x - (14 - x^{2}) d x + \int_{- \frac{14}{3}}^{0} \frac{5}{3} x - 5 x d x + \int_{0}^{2} 5 x - \frac{5}{3} x d x + \int_{2}^{3} \frac{5}{3} x - (14 - x^{2}) d x

\int_{- 7}^{- \frac{14}{3}} 5 x - (14 - x^{2}) d x = \frac{3283}{162}

\int_{- \frac{14}{3}}^{0} \frac{5}{3} x - 5 x d x = \frac{980}{27}

\int_{0}^{2} 5 x - \frac{5}{3} x d x = \frac{20}{3}

\int_{2}^{3} \frac{5}{3} x - (14 - x^{2}) d x = \frac{7}{2}

= \frac{3283}{162} + \frac{980}{27} + \frac{20}{3} + \frac{7}{2}

Simplify

= \frac{5405}{81}

9 e^{x}

y (X) = \frac{1}{2 cos ^{2} ( X )}

y = 4 x ln (x) + (3^{x}) - (x^{3})

\int \frac{1}{( 256 + x ^{2} ) ^{\frac{3}{2}}} d x

x \to 0 lim ((3 x^{2} - 2)^{5})