$\int_{0}^{15} x^{2} \sqrt{a^{2} - x^{2}} d x$ $int_(0)^(15)x^2sqrt(a^2-x^2)dx$ ?

Antworten:

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x = \frac{a^{4} π}{16}$ $int_0^a x^2sqrt(a^2-x^2)dx = (a^4pi)/16$

Erläuterung:

Bewerten:

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x$ $int_0^a x^2sqrt(a^2-x^2)dx$

Ersatz:

$x = a sin t$ $x= asint$

$d x = a cos t d t$ $dx = a costdt$

mit $t \in [0, \frac{π}{2}]$ $t in [0,pi/2]$

damit:

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x = \int_{0}^{\frac{π}{2}} a^{2} {sin}^{2} t \sqrt{a^{2} - a^{2} {sin}^{2} t} a cos t d t$ $int_0^a x^2sqrt(a^2-x^2)dx = int_0^(pi/2) a^2 sin^2t sqrt(a^2-a^2 sin^2t)acostdt$

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x = a^{4} \int_{0}^{\frac{π}{2}} {sin}^{2} t \sqrt{1 - {sin}^{2} t} cos t d t$ $int_0^a x^2sqrt(a^2-x^2)dx = a^4 int_0^(pi/2) sin^2t sqrt(1- sin^2t)costdt$

Für $t \in [0, \frac{π}{2}]$ $t in [0,pi/2]$ Der Cosinus ist positiv, also:

$\sqrt{1 - {sin}^{2} t} = cos t$ $sqrt(1- sin^2t) = cost$

und dann:

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x = a^{4} \int_{0}^{\frac{π}{2}} {sin}^{2} t {cos}^{2} t d t$ $int_0^a x^2sqrt(a^2-x^2)dx = a^4 int_0^(pi/2) sin^2t cos^2tdt$

Verwenden Sie jetzt die trigonometrischen Identitäten:

$sin 2 θ = 2 sin θ cos θ$ $sin 2theta = 2 sin theta cos theta$

$2 {sin}^{2} θ = 1 - cos θ$ $2sin^2 theta = 1- cos theta$

so:

${sin}^{2} t {cos}^{2} t d t = \frac{1}{4} {sin}^{2} 2 t = \frac{1}{8} (1 - cos 4 t)$ $sin^2t cos^2tdt = 1/4 sin^2 2t = 1/8(1-cos4t)$

und:

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x = \frac{a^{4}}{8} \int_{0}^{\frac{π}{2}} (1 - cos 4 t) d t$ $int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 int_0^(pi/2) (1-cos4t)dt$

Verwenden Sie jetzt die Linearität des Integrals:

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x = \frac{a^{4}}{8} (\int_{0}^{\frac{π}{2}} d t - \int_{0}^{\frac{π}{2}} cos 4 t d t)$ $int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 (int_0^(pi/2) dt - int_0^(pi/2) cos4tdt)$

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x = \frac{a^{4}}{8} {[t - \frac{sin 4 t}{4}]}_{0}^{\frac{π}{2}}$ $int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 [t - (sin 4t)/4]_0^(pi/2)$

$\int_{0}^{a} x^{2} \sqrt{a^{2} - x^{2}} d x = \frac{a^{4} π}{16}$ $int_0^a x^2sqrt(a^2-x^2)dx = (a^4pi)/16$

int_(0)^(15)x^2sqrt(a^2-x^2)dx∫150x2√a2−x2dxint_(0)^(15)x^2sqrt(a^2-x^2)dx?

Antworten:

Erläuterung:

$\int_{0}^{15} x^{2} \sqrt{a^{2} - x^{2}} d x$ $int_(0)^(15)x^2sqrt(a^2-x^2)dx$ ?