What is area x^2+8x-3,-x^2+4x+3 ?

The answer to area x^2+8x-3,-x^2+4x+3 is 64/3

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area x^2+8x-3,-x^2+4x+3

area

x^{2} + 8 x - 3, - x^{2} + 4 x + 3

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

Apply the area formula:

\int_{- 3}^{1} x^{2} + 8 x - 3 - (- x^{2} + 4 x + 3) d x

Solve

\int_{- 3}^{1} x^{2} + 8 x - 3 - (- x^{2} + 4 x + 3) d x : \frac{64}{3}

The area is:

= \frac{64}{3}

n = 1 \sum \infty \frac{5 ^{n}}{9}

f (x) = \frac{sin ( 4 x )}{sin ( 5 x )}

y = arctan (3 x^{2})

\int e^{10 t} d t

\frac{d}{d x} (f (x)^{- 1} (x))