integral from 1 to 2 of (3t^2+2e^{3t}+1/t+2cos(3t))

Lösung

\int_{1}^{2} (3 t^{2} + 2 e^{3 t} + \frac{1}{t} + 2 cos (3 t)) d t

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

Dezimale

262.97496 \dots

Schritte zur Lösung

\int_{1}^{2} 3 t^{2} + 2 e^{3 t} + \frac{1}{t} + 2 cos (3 t) d t

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}^{2} 3 t^{2} d t + \int_{1}^{2} 2 e^{3 t} d t + \int_{1}^{2} \frac{1}{t} d t + \int_{1}^{2} 2 cos (3 t) d t

\int_{1}^{2} 3 t^{2} d t = 7

\int_{1}^{2} 2 e^{3 t} d t = \frac{2 ( e ^{6} - e ^{3} )}{3}

\int_{1}^{2} \frac{1}{t} d t = ln (2)

\int_{1}^{2} 2 cos (3 t) d t = \frac{2}{3} (sin (6) - sin (3))

= 7 + \frac{2 ( e ^{6} - e ^{3} )}{3} + ln (2) + \frac{2}{3} (sin (6) - sin (3))

Vereinfache

7 + \frac{2 ( e ^{6} - e ^{3} )}{3} + ln (2) + \frac{2}{3} (sin (6) - sin (3)) : ln (2) + \frac{2 ( e ^{6} - e ^{3} ) + 2 ( sin ( 6 ) - sin ( 3 ) )}{3} + 7

= ln (2) + \frac{2 ( e ^{6} - e ^{3} ) + 2 ( sin ( 6 ) - sin ( 3 ) )}{3} + 7

\int_{b}^{a} 1 d x

\int_{0}^{4} \frac{1}{( 2 - 3 x ) ^{\frac{1}{3}}} d x

\int_{0}^{x^{2}} e^{x^{3}} d x

\int_{4}^{9} 5 x d x

\int_{1}^{x^{2}} f (t) d t