integral from 5 to 10 of 1/(36-9x^2)

Lösung

\int_{5}^{10} \frac{1}{36 - 9 x ^{2}} d x

\frac{2 ln ( \frac{3}{2} ) - ln ( \frac{7}{2} )}{36}

Dezimale

- 0.01227 \dots

Schritte zur Lösung

\int_{5}^{10} \frac{1}{36 - 9 x ^{2}} d x

Wende integrale Substitution an

= \int_{\frac{5}{2}}^{5} \frac{1}{18 ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{18} \cdot \int_{\frac{5}{2}}^{5} \frac{1}{- u ^{2} + 1} d u

Nutze das gemeinsame Integral :

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{18} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{5}{2}}^{5}

Vereinfache

\frac{1}{18} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{5}{2}}^{5} : \frac{1}{18} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{5}{2}}^{5}

= \frac{1}{18} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{5}{2}}^{5}

Berechne die Grenzen:

\frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( \frac{7}{2} ) + ln ( \frac{3}{2} )}{2}

= \frac{1}{18} \cdot \frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( \frac{7}{2} ) + ln ( \frac{3}{2} )}{2}

Vereinfache

= \frac{2 ln ( \frac{3}{2} ) - ln ( \frac{7}{2} )}{36}

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

\int_{- \infty}^{- 2} 3 e^{x} d x

\int_{5}^{\infty} \frac{5}{x} d x

\int_{- 1}^{1} 2 x^{2} - x^{3} d x

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