What is the integral from 0 to x of (t-1)^2(t-3)^3 ?

The integral from 0 to x of (t-1)^2(t-3)^3 is (x^6)/6-(11x^5)/5+(23x^4)/2-30x^3+(81x^2)/2-27x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to x of (t-1)^2(t-3)^3

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

\frac{x ^{6}}{6} - \frac{11 x ^{5}}{5} + \frac{23 x ^{4}}{2} - 30 x^{3} + \frac{81 x ^{2}}{2} - 27 x

Solution steps

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

Expand

(t - 1)^{2} (t - 3)^{3} : t^{5} - 11 t^{4} + 46 t^{3} - 90 t^{2} + 81 t - 27

= \int_{0}^{x} t^{5} - 11 t^{4} + 46 t^{3} - 90 t^{2} + 81 t - 27 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{x} t^{5} d t - \int_{0}^{x} 11 t^{4} d t + \int_{0}^{x} 46 t^{3} d t - \int_{0}^{x} 90 t^{2} d t + \int_{0}^{x} 81 t d t - \int_{0}^{x} 27 d t

\int_{0}^{x} t^{5} d t = \frac{x ^{6}}{6}

\int_{0}^{x} 11 t^{4} d t = \frac{11 x ^{5}}{5}

\int_{0}^{x} 46 t^{3} d t = \frac{23 x ^{4}}{2}

\int_{0}^{x} 90 t^{2} d t = 30 x^{3}

\int_{0}^{x} 81 t d t = \frac{81 x ^{2}}{2}

\int_{0}^{x} 27 d t = 27 x

= \frac{x ^{6}}{6} - \frac{11 x ^{5}}{5} + \frac{23 x ^{4}}{2} - 30 x^{3} + \frac{81 x ^{2}}{2} - 27 x

\int_{0}^{0.2} \frac{x ^{6}}{1 + x ^{7}} d x

\int_{0}^{t} cos^{2} (x) d x

\int_{0}^{4} x - 3 d x

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

\int_{7}^{x} sin (t^{4}) d t

What is the integral from 0 to x of (t-1)^2(t-3)^3 ?
The integral from 0 to x of (t-1)^2(t-3)^3 is (x^6)/6-(11x^5)/5+(23x^4)/2-30x^3+(81x^2)/2-27x