What is the integral from 1 to 2 of (-y^3+2y^2+5y-6) ?

The integral from 1 to 2 of (-y^3+2y^2+5y-6) is 29/12

Popular >

integral from 1 to 2 of (-y^3+2y^2+5y-6)

\int_{1}^{2} (- y^{3} + 2 y^{2} + 5 y - 6) d y

\frac{29}{12}

Decimal Notation

2.41666 \dots

Solution steps

\int_{1}^{2} - y^{3} + 2 y^{2} + 5 y - 6 d y

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}^{2} y^{3} d y + \int_{1}^{2} 2 y^{2} d y + \int_{1}^{2} 5 y d y - \int_{1}^{2} 6 d y

\int_{1}^{2} y^{3} d y = \frac{15}{4}

\int_{1}^{2} 2 y^{2} d y = \frac{14}{3}

\int_{1}^{2} 5 y d y = \frac{15}{2}

\int_{1}^{2} 6 d y = 6

= - \frac{15}{4} + \frac{14}{3} + \frac{15}{2} - 6

Simplify

- \frac{15}{4} + \frac{14}{3} + \frac{15}{2} - 6 : \frac{29}{12}

= \frac{29}{12}

\int_{- 3}^{3} ∣ 2 x - 1 ∣ d x

11 + \frac{1}{5} sin (11 π t)

\int 3 x^{2} x^{3} + 1 d x

\int (sin (x) - 8 cos (x)) d x

\int_{0}^{1} 4 x^{3} e^{x^{4}} d x

What is the integral from 1 to 2 of (-y^3+2y^2+5y-6) ?
The integral from 1 to 2 of (-y^3+2y^2+5y-6) is 29/12