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

Lösung

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

2

Schritte zur Lösung

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

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

\int_{1}^{2} 4 x d x = 6

\int_{1}^{2} 1 d x = 1

= 7 - 6 + 1

Vereinfache

= 2

\int_{- \infty}^{- 3} \frac{4}{x ^{3}} d x

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\int_{0}^{2} (x^{2} + 3 x - 1) e^{2 x} d x

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

\int_{\frac{π}{4}}^{\frac{π}{2}} 7 cot^{3} (x) d x