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

Lösung

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

\frac{184}{15}

Dezimale

12.26666 \dots

Schritte zur Lösung

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

Wende die Summenregel an:

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

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

\int_{1}^{3} (4 x - 3)^{2} d x = \frac{182}{3}

\int_{1}^{3} (x^{2})^{2} d x = \frac{242}{5}

= \frac{182}{3} - \frac{242}{5}

Vereinfache

\frac{182}{3} - \frac{242}{5} : \frac{184}{15}

= \frac{184}{15}

\int \frac{6 x ^{2} + 8}{x ^{3} + 4 x} d x

\frac{d y}{d x} + y = e^{7 x}

y^{^{'}} - \frac{5}{x} y = \frac{y ^{5}}{x ^{18}}

\int_{9}^{16} \frac{1}{7 - x} d x

\int \frac{x}{x - 16} d x