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 16/3

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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{16}{3}

Decimal Notation

5.33333 \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{28}{3}

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

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

= \frac{28}{3} - 16 + 12

Simplify

\frac{28}{3} - 16 + 12 : \frac{16}{3}

= \frac{16}{3}

\int sin^{2} (2 x) cos^{2} (2 x) d x

\frac{d}{d x} (\frac{3 x ^{2} + 5 x + 7}{x})

\int_{0}^{\frac{π}{12}} \frac{sin ( 6 x )}{1 + cos ^{2} ( 6 x )} d x

3 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

\frac{d}{d x} (2 sin (x^{2} - 1))

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 16/3