integral from 0 to 1 of sqrt(1+(9x)/4)

Lösung

\int_{0}^{1} 1 + \frac{9 x}{4} d x

\frac{13 13 - 8}{27}

Dezimale

1.43970 \dots

Schritte zur Lösung

\int_{0}^{1} 1 + \frac{9 x}{4} d x

Wende U-Substitution an

= \int_{1}^{\frac{13}{4}} \frac{4 u}{9} d u

Entferne die Konstante:

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

= \frac{4}{9} \cdot \int_{1}^{\frac{13}{4}} u d u

Wende die Potenzregel an

= \frac{4}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{13}{4}}

Berechne die Grenzen:

\frac{13 13}{12} - \frac{2}{3}

= \frac{4}{9} (\frac{13 13}{12} - \frac{2}{3})

Vereinfache

= \frac{13 13 - 8}{27}

cos (t) y^{^{'}} + sin (t) y = 2 cos^{3} (t) sin (t) - 1

d er i v a t i v e \frac{c}{x}

d er i v a t i v e 2^{(x^{2} - 3 x + 8)}

t an g e n t f (x) = 5 + 8 x, at x = 0

x \to \infty lim (x \cdot \frac{1}{x})