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integral from 1/12 to 1/4 of 2/(sqrt(t)+4t\sqrt{t)}

Lösung

\int_{\frac{1}{12}}^{\frac{1}{4}} \frac{2}{t + 4 t t} d t

\frac{π}{6}

Dezimale

0.52359 \dots

Schritte zur Lösung

\int_{\frac{1}{12}}^{\frac{1}{4}} \frac{2}{t + 4 t t} d t

Entferne die Konstante:

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

= 2 \cdot \int_{\frac{1}{12}}^{\frac{1}{4}} \frac{1}{t + 4 t t} d t

\frac{1}{t + 4 t t} = \frac{t - 4 t t}{( t ) ^{2} - ( 4 t t ) ^{2}}

= 2 \cdot \int_{\frac{1}{12}}^{\frac{1}{4}} \frac{t - 4 t t}{( t ) ^{2} - ( 4 t t ) ^{2}} d t

Vereinfache

\frac{t - 4 t t}{( t ) ^{2} - ( 4 t t ) ^{2}} : \frac{1}{t ( 4 t + 1 )}

= 2 \cdot \int_{\frac{1}{12}}^{\frac{1}{4}} \frac{1}{t ( 4 t + 1 )} d t

Wende U-Substitution an

= 2 \cdot \int_{\frac{1}{2 3}}^{\frac{1}{2}} \frac{2}{4 u ^{2} + 1} d u

Entferne die Konstante:

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

= 2 \cdot 2 \cdot \int_{\frac{1}{2 3}}^{\frac{1}{2}} \frac{1}{4 u ^{2} + 1} d u

Wende integrale Substitution an

= 2 \cdot 2 \cdot \int_{\frac{1}{3}}^{1} \frac{1}{2 ( v ^{2} + 1 )} d v

Entferne die Konstante:

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

= 2 \cdot 2 \cdot \frac{1}{2} \cdot \int_{\frac{1}{3}}^{1} \frac{1}{v ^{2} + 1} d v

Nutze das gemeinsame Integral :

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot 2 \cdot \frac{1}{2} [arctan (v)]_{\frac{1}{3}}^{1}

Vereinfache

2 \cdot 2 \cdot \frac{1}{2} [arctan (v)]_{\frac{1}{3}}^{1} : 2 [arctan (v)]_{\frac{1}{3}}^{1}

= 2 [arctan (v)]_{\frac{1}{3}}^{1}

Berechne die Grenzen:

\frac{π}{12}

= 2 \cdot \frac{π}{12}

Vereinfache

= \frac{π}{6}

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

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