What is area |4x-1|,x^2-5 ?

The answer to area |4x-1|,x^2-5 is (371+80sqrt(10)+128sqrt(2))/(12)

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area |4x-1|,x^2-5

area

∣ 4 x - 1 ∣, x^{2} - 5

\frac{371 + 80 10 + 128 2}{12}

Decimal Notation

67.08346 \dots

Solution steps

Apply the area formula:

\int_{- 2 - 10}^{2 (1 + 2)} ∣ 4 x - 1 ∣ - (x^{2} - 5) d x

Solve

\int_{- 2 - 10}^{2 (1 + 2)} ∣ 4 x - 1 ∣ - (x^{2} - 5) d x : \frac{371 + 80 10 + 128 2}{12}

The area is:

= \frac{371 + 80 10 + 128 2}{12}

\frac{d y}{d x} = - (\frac{y ^{2}}{1 + y ^{2}} e^{(- x)})

x \to 4 lim (7)

x \to 0 lim (\frac{1}{8} + x - (\frac{1}{8}))

\int (5 x^{2} + 3 x - 10) d x

\int \frac{18 x ^{2}}{x ^{4} - 41 x ^{2} + 400} d x

What is area |4x-1|,x^2-5 ?
The answer to area |4x-1|,x^2-5 is (371+80sqrt(10)+128sqrt(2))/(12)