What is the integral from-1 to 4 of 4-x^2+3x ?

The integral from-1 to 4 of 4-x^2+3x is 125/6

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integral from-1 to 4 of 4-x^2+3x

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

\frac{125}{6}

Decimal Notation

20.83333 \dots

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{4} 4 d x - \int_{- 1}^{4} x^{2} d x + \int_{- 1}^{4} 3 x d x

\int_{- 1}^{4} 4 d x = 20

\int_{- 1}^{4} x^{2} d x = \frac{65}{3}

\int_{- 1}^{4} 3 x d x = \frac{45}{2}

= 20 - \frac{65}{3} + \frac{45}{2}

Simplify

20 - \frac{65}{3} + \frac{45}{2} : \frac{125}{6}

= \frac{125}{6}

\int (\frac{1}{3}) x^{3} d x

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

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

\frac{d}{d x} (\frac{- x ^{2} - 3}{( x ^{2} - 3 ) ^{2}})

6 x^{2} + 3 x y + y^{2} = 16, (1, 2)

What is the integral from-1 to 4 of 4-x^2+3x ?
The integral from-1 to 4 of 4-x^2+3x is 125/6