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

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

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

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

3

Solution steps

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

Factor

x^{2} + x + \frac{1}{4} : (x + 0.5)^{2}

= \int_{- 3}^{- 1} (x + 0.5)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{- 3}^{- 1} (x + 0.5)^{2} d x

Eliminate Absolutes

= \int_{- 3}^{- 1} (x + 0.5)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{- 3}^{- 1} ∣ (x + 0.5) ∣ d x

Eliminate Absolutes

= \int_{- 3}^{- 1} - x - 0.5 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_{- 3}^{- 1} x d x - \int_{- 3}^{- 1} 0.5 d x

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

\int_{- 3}^{- 1} 0.5 d x = 1

= - (- 4) - 1

Simplify

= 3

\int_{0}^{\infty} e^{- 66 x} x d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} - cos (x) d x

\int_{0}^{1} 1 + x d x

\int_{0}^{2} \frac{x ^{2}}{2 + x ^{3}} d x

\int_{0}^{2} (x^{3} + 1) d x

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