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

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

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

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 4 x d x = 16

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

= - \frac{26}{3} + 16 - 6

Simplify

- \frac{26}{3} + 16 - 6 : \frac{4}{3}

= \frac{4}{3}

x \to (0) lim (\frac{ln ( x )}{x})

x \to - 3 lim (2 x - 3 a x + 6 a)

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

\frac{\partial}{\partial x} (\frac{- x}{y ^{2} + x ^{2}})

\int \frac{( x ^{3} - 4 )}{x ^{3} + 2 x ^{2} + 2 x} d x

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