What is area y=x^2+1,y=x+3,(-1,2) ?

The answer to area y=x^2+1,y=x+3,(-1,2) is 9/2

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area y=x^2+1,y=x+3,(-1,2)

area

y = x^{2} + 1, y = x + 3, (- 1, 2)

\frac{9}{2}

Decimal Notation

4.5

Solution steps

Apply the area formula:

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

Solve

\int_{- 1}^{2} x^{2} + 1 - (x + 3) d x : \frac{9}{2}

The area is:

= \frac{9}{2}

\frac{1}{s ( s + 1 )}

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e^{x^{2}} \cdot \frac{1}{2}

q^{x} + x^{q}

\int_{5}^{- 3} 10 x^{2} d x

What is area y=x^2+1,y=x+3,(-1,2) ?
The answer to area y=x^2+1,y=x+3,(-1,2) is 9/2