What is area-x^2+3x+1,-x-1,[0,4] ?

The answer to area-x^2+3x+1,-x-1,[0,4] is 56/3

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area-x^2+3x+1,-x-1,[0,4]

area

- x^{2} + 3 x + 1, - x - 1, [0, 4]

\frac{56}{3}

Decimal Notation

18.66666 \dots

Solution steps

Apply the area formula:

\int_{0}^{4} - x^{2} + 3 x + 1 - (- x - 1) d x

Solve

\int_{0}^{4} - x^{2} + 3 x + 1 - (- x - 1) d x : \frac{56}{3}

The area is:

= \frac{56}{3}

\int \frac{- 3}{x} d x

\int \frac{4 t - 3}{t + 1} d t

f (x) = x^{4} + 5 x^{2} - x, (1, 5)

\int \frac{4 x + 3}{x ^{2} + 4} d x

(x^{4} - x + y) d x - x d y = 0

What is area-x^2+3x+1,-x-1,[0,4] ?
The answer to area-x^2+3x+1,-x-1,[0,4] is 56/3