integral from y to y+1 of 2y

Lösung

\int_{y}^{y + 1} 2 y d y

2 y + 1

Schritte zur Lösung

\int_{y}^{y + 1} 2 y d y

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{y}^{y + 1} y d y

Wende die Potenzregel an

= 2 [\frac{y ^{2}}{2}]_{y}^{y + 1}

Berechne die Grenzen:

\frac{( y + 1 ) ^{2}}{2} - \frac{y ^{2}}{2}

= 2 (\frac{( y + 1 ) ^{2}}{2} - \frac{y ^{2}}{2})

Vereinfache

= 2 y + 1

\int_{0}^{12} x (\frac{1}{72} x) d x

\int_{1}^{7} 6 x^{2} + 2 x - 3 d x

\int_{0}^{2} 1 - 0.5 x d x

\int_{0}^{3} \frac{- 2}{5 x + 1} d x

\int_{- 1}^{2} 4 t^{2} - 8 t + 5 d t