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

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

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integral from 1 to 3 of (3-x^2)-(6-4x)

\int_{1}^{3} (3 - x^{2}) - (6 - 4 x) d x

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

\int_{1}^{3} 3 - x^{2} - (6 - 4 x) d x

Simplify

= \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}

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

\int 7 x + 9 d x

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

e^{2 x^{3}}

\frac{d}{d x} (3 sec (x) tan (x))

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