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

Lösung

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

ln ∣ 4 - x ∣ - ln ∣ x ∣ + 4 x^{2} (- x + 2)

Schritte zur Lösung

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{x}^{4 - x} \frac{1}{y} d y + \int_{x}^{4 - x} x^{2} y d y

\int_{x}^{4 - x} \frac{1}{y} d y = ln ∣ 4 - x ∣ - ln ∣ x ∣

\int_{x}^{4 - x} x^{2} y d y = 4 x^{2} (- x + 2)

= ln ∣ 4 - x ∣ - ln ∣ x ∣ + 4 x^{2} (- x + 2)

\int_{- 4}^{3} ∣ x ∣ d x

\int_{0}^{1} e^{x + 1} d x

\int_{2}^{5} (2 x - 1) d x

\int_{0}^{1} x^{p} d x

\int_{3}^{6} 30 x^{3} 1 + x^{2} d x