Let us determine the points of intersection of y = x 2 y=x^2 y = x 2 and y = 2 x + 3 y=2x+3 y = 2 x + 3 :
x 2 = 2 x + 3 , x 2 − 2 x − 3 = 0 , D = ( − 2 ) 2 − 4 ⋅ 1 ⋅ ( − 3 ) = 16 = 4 2 , x 1 , 2 = 2 ± 4 2 = 3 o r − 1. x^2 = 2x + 3, \\
x^2 - 2x-3 = 0, \\
D = (-2)^2 - 4\cdot 1\cdot (-3) = 16=4^2, \\
x_{1,2} = \dfrac{2\pm 4}{2} = 3\,\,\,\mathrm{or} \,\, -1. x 2 = 2 x + 3 , x 2 − 2 x − 3 = 0 , D = ( − 2 ) 2 − 4 ⋅ 1 ⋅ ( − 3 ) = 16 = 4 2 , x 1 , 2 = 2 2 ± 4 = 3 or − 1.
So we need to evaluate the integral equal to the difference of areas under the graphs when x ∈ [ − 1 ; 3 ] x\in[-1;3] x ∈ [ − 1 ; 3 ]
S = ∫ − 1 3 ( 2 x + 3 − x 2 ) d x = ( x 2 + 3 x − x 3 3 ) ∣ − 1 3 , S = ( 3 2 + 3 ⋅ 3 − 3 3 3 ) − ( ( − 1 ) 2 + 3 ⋅ ( − 1 ) − ( − 1 ) 3 3 ) , S = 32 3 . S=\int\limits_{-1}^3 \left(2x+3 - x^2 \right)\,dx = \left( x^2 + 3x - \dfrac{x^3}{3}\right) \Big|_{-1}^3 \,, \\
S = \left(3^2 + 3\cdot3 - \dfrac{3^3}{3} \right) - \left( (-1)^2 + 3\cdot(-1) - \dfrac{(-1)^3}{3}\right), \\
S = \dfrac{32}{3}. S = − 1 ∫ 3 ( 2 x + 3 − x 2 ) d x = ( x 2 + 3 x − 3 x 3 ) ∣ ∣ − 1 3 , S = ( 3 2 + 3 ⋅ 3 − 3 3 3 ) − ( ( − 1 ) 2 + 3 ⋅ ( − 1 ) − 3 ( − 1 ) 3 ) , S = 3 32 .
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