What is the integral from-1 to 3 of (x^2+6x+3) ?

The integral from-1 to 3 of (x^2+6x+3) is 136/3

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integral from-1 to 3 of (x^2+6x+3)

\int_{- 1}^{3} (x^{2} + 6 x + 3) d x

\frac{136}{3}

Decimal Notation

45.33333 \dots

Solution steps

\int_{- 1}^{3} x^{2} + 6 x + 3 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}^{3} x^{2} d x + \int_{- 1}^{3} 6 x d x + \int_{- 1}^{3} 3 d x

\int_{- 1}^{3} x^{2} d x = \frac{28}{3}

\int_{- 1}^{3} 6 x d x = 24

\int_{- 1}^{3} 3 d x = 12

= \frac{28}{3} + 24 + 12

Simplify

\frac{28}{3} + 24 + 12 : \frac{136}{3}

= \frac{136}{3}

\frac{d ^{3}}{d x ^{3}} (x^{4} - 10 x^{3})

\frac{\partial}{\partial x} (\frac{1}{y + z})

f (x) = \frac{5}{x ^{5}}

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} + 9 e^{5 t}

n \to \infty lim (\frac{i}{n})

What is the integral from-1 to 3 of (x^2+6x+3) ?
The integral from-1 to 3 of (x^2+6x+3) is 136/3