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

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

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integral from-1 to 4 of 3x-(x^2-4)

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

\frac{125}{6}

Decimal Notation

20.83333 \dots

Solution steps

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

Simplify

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

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

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

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

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

Simplify

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

= \frac{125}{6}

f (x) = - x^{3}, at x = 1

\int 9 x^{4} (x^{5} + 2)^{8} d x

- 2 e^{- x^{2}}

y = \frac{csc ( 4 x )}{ln ( x )}

x^{2} \frac{d ^{2} y}{d x ^{2}} + 4 x \frac{d y}{d x} + 2 y = 8 x

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