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

The integral from 0 to 2 of (2-t)^2t^3 is 16/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{4} t^{4} (2 - t)^{2} - \int - \frac{1}{2} t^{4} (- t + 2) d t]_{0}^{2}

= [\frac{1}{4} t^{4} (2 - t)^{2} - \int - \frac{1}{2} t^{4} (- t + 2) d t]_{0}^{2}

\int - \frac{1}{2} t^{4} (- t + 2) d t = \frac{t ^{6}}{12} - \frac{t ^{5}}{5}

= [\frac{1}{4} t^{4} (2 - t)^{2} - (\frac{t ^{6}}{12} - \frac{t ^{5}}{5})]_{0}^{2}

Simplify

[\frac{1}{4} t^{4} (2 - t)^{2} - (\frac{t ^{6}}{12} - \frac{t ^{5}}{5})]_{0}^{2} : [\frac{1}{30} t^{4} (5 t^{2} - 24 t + 30)]_{0}^{2}

= [\frac{1}{30} t^{4} (5 t^{2} - 24 t + 30)]_{0}^{2}

Compute the boundaries:

\frac{16}{15}

= \frac{16}{15}

\int_{0}^{2} - x^{2} + 4 x - (x^{2}) d x

\int_{4}^{5} \frac{1}{x ^{2} - 5 x + 6} d x

\int_{2}^{14} \frac{1}{x} d x

\int_{0}^{4} x^{3} + x - x^{2} d x

\int_{\frac{π}{4}}^{\frac{π}{2}} 4 cos^{4} (x) d x

What is the integral from 0 to 2 of (2-t)^2t^3 ?
The integral from 0 to 2 of (2-t)^2t^3 is 16/15