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

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}

s l o p e 6 x^{2} + 3 x y + y^{2} = 16, (1, 2)

\frac{d}{d x} (2 d x)

\int \frac{1}{e ^{2 x} + 4} d x

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{- x}

\int (cos (x)) e^{s i n (x)} d x

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