Propriété
Soit f f f [ a ; b ] [a ; b] [ a ; b ] ( C f ) (C_f) ( C f ) ( O , u ⃗ , v ⃗ ) (O,\vec{u},\vec{v}) ( O , u , v )
L’aire, exprimée en unités d’aire, du domaine situé entre la courbe ( C f ) (C_f) ( C f ) x = a x = a x = a x = b x = b x = b
A = ∫ a b ∣ f ( x ) ∣ d x . ∥ i ⃗ ∥ . ∥ j ⃗ ∥ \mathcal{A}=\int_{a}^{b}\left|f(x)\right|dx .\|\vec{i}\|.\|\vec{j}\| A = ∫ a b ∣ f ( x ) ∣ d x .∥ i ∥.∥ j ∥ Remarques :
si f f f [ a ; b ] [a ; b] [ a ; b ] [ a , b ] [a,b] [ a , b ]
A = ∫ a b f ( x ) d x . . ∥ i ⃗ ∥ . ∥ j ⃗ ∥ \mathcal{A}=\int_{a}^{b}f(x)dx ..\|\vec{i}\|.\|\vec{j}\| A = ∫ a b f ( x ) d x ..∥ i ∥.∥ j ∥
si f f f [ a ; b ] [a ; b] [ a ; b ]
A = − ∫ a b f ( x ) d x . . ∥ i ⃗ ∥ . ∥ j ⃗ ∥ \mathcal{A}=-\int_{a}^{b}f(x)dx ..\|\vec{i}\|.\|\vec{j}\| A = − ∫ a b f ( x ) d x ..∥ i ∥.∥ j ∥
Application 6
Soit f f f f ( x ) = − x 2 + 2 x f(x)=-x^2+2x f ( x ) = − x 2 + 2 x
( C f ) (C_f) ( C f ) ( O , i ⃗ , j ⃗ ) (O,\vec{i},\vec{j}) ( O , i , j ) ∥ i ⃗ ∥ = 2 c m \|\vec{i}\|=2cm ∥ i ∥ = 2 c m ∥ j ⃗ ∥ = 2 c m \|\vec{j}\|=2cm ∥ j ∥ = 2 c m ( C f ) (C_f) ( C f ) x = 0 x = 0 x = 0 x = 3 x = 3 x = 3
Correction
L’aire du domaine situé entre la courbe
( C f ) (C_f) ( C f ) x = 0 x = 0 x = 0 x = 3 x = 3 x = 3
A = ∫ 0 3 ∣ f ( x ) ∣ d x . ∥ i ⃗ ∥ . ∥ j ⃗ ∥ = 4 ∫ 0 3 ∣ f ( x ) ∣ d x ⋅ c m 2 \mathcal{A}=\int_{0}^{3}\left|f(x)\right|dx .\|\vec{i}\|.\|\vec{j}\|
=4\int_{0}^{3}\left|f(x)\right|dx\cdot cm^2 A = ∫ 0 3 ∣ f ( x ) ∣ d x .∥ i ∥.∥ j ∥ = 4 ∫ 0 3 ∣ f ( x ) ∣ d x ⋅ c m 2 ∫ 0 3 ∣ f ( x ) ∣ d x = ∫ 0 2 ∣ f ( x ) ∣ d x + ∫ 2 3 ∣ f ( x ) ∣ d x = ∫ 0 2 f ( x ) d x − ∫ 2 3 f ( x ) d x = ∫ 0 2 ( − x 2 + 2 x ) d x − ∫ 2 3 ( − x 2 + 2 x ) d x = [ − x 3 3 + x 2 ] 0 2 − [ − x 3 3 + x 2 ] 2 3 = − 8 3 + 4 − ( − 9 + 9 + 8 3 − 4 ) = 8 3 \begin{align*}
\int_{0}^{3}\left|f(x)\right|dx
&= \int_{0}^{2}\left|f(x)\right|dx + \int_{2}^{3}\left|f(x)\right|dx \\
&= \int_{0}^{2}f(x)dx - \int_{2}^{3}f(x)dx \\
&= \int_{0}^{2}(-x^2+2x)dx - \int_{2}^{3}(-x^2+2x)dx \\
&=\left[-\frac{x^3}3+x^2\right]^2_0 - \left[-\frac{x^3}3+x^2\right]^3_2 \\
&=-\frac83+4-(-9+9+\frac83-4) \\
&=\frac{8}3
\end{align*} ∫ 0 3 ∣ f ( x ) ∣ d x = ∫ 0 2 ∣ f ( x ) ∣ d x + ∫ 2 3 ∣ f ( x ) ∣ d x = ∫ 0 2 f ( x ) d x − ∫ 2 3 f ( x ) d x = ∫ 0 2 ( − x 2 + 2 x ) d x − ∫ 2 3 ( − x 2 + 2 x ) d x = [ − 3 x 3 + x 2 ] 0 2 − [ − 3 x 3 + x 2 ] 2 3 = − 3 8 + 4 − ( − 9 + 9 + 3 8 − 4 ) = 3 8 A = 32 3 c m 2 \mathcal{A}=\frac{32}3cm^2 A = 3 32 c m 2
Propriété
Soient f f f g g g [ a , b ] [a,b] [ a , b ] ( C f ) (C_f) ( C f ) ( C g ) (C_g) ( C g ) f f f g g g
L’aire de la surface située entre les courbes de f f f g g g x = a x=a x = a x = b x=b x = b
A = ∫ a b ∣ f ( x ) − g ( x ) ∣ d x . ∥ i ⃗ ∥ . ∥ j ⃗ ∥ \mathscr{A}=\int_{a}^{b}\left|f(x)-g(x)\right|dx .\|\vec{i}\|.\|\vec{j}\| A = ∫ a b ∣ f ( x ) − g ( x ) ∣ d x .∥ i ∥.∥ j ∥ Exemple :
f ( x ) = c o s ( x ) f(x)=cos(x) f ( x ) = cos ( x ) g ( x ) = s i n ( x ) g(x)=sin(x) g ( x ) = s in ( x )
( O , i ⃗ , j ⃗ ) (O,\vec{i},\vec{j}) ( O , i , j ) ∥ i ⃗ ∥ = ∥ j ⃗ ∥ = 1 c m \|\vec{i}\|=\|\vec{j}\|=1cm ∥ i ∥ = ∥ j ∥ = 1 c m
Calculer l’aire du domaine situé entre la courbe
( C f ) (C_f) ( C f ) ( C g ) (C_g) ( C g ) x = − π 2 x = -\frac\pi2 x = − 2 π x = 3 π 4 x = \frac{3\pi}{4} x = 4 3 π
L’aire du domaine situé entre la courbe
( C f ) (C_f) ( C f ) ( C g ) (C_g) ( C g ) x = − π 2 x = -\frac\pi2 x = − 2 π x = 3 π 4 x = \frac{3\pi}{4} x = 4 3 π
A = ∫ − π 2 3 π 4 ∣ f ( x ) − g ( x ) ∣ d x . ∥ i ⃗ ∥ . ∥ j ⃗ ∥ = ∫ − π 2 3 π 4 ∣ f ( x ) − g ( x ) ∣ d x ⋅ c m 2 \begin{align*}
\mathcal{A}&=\int_{-\frac\pi2}^{\frac{3\pi}{4}}\left|f(x)-g(x)\right|dx .\|\vec{i}\|.\|\vec{j}\|\\
&=\int_{-\frac\pi2}^{\frac{3\pi}{4}}\left|f(x)-g(x)\right|dx\cdot cm^2
\end{align*} A = ∫ − 2 π 4 3 π ∣ f ( x ) − g ( x ) ∣ d x .∥ i ∥.∥ j ∥ = ∫ − 2 π 4 3 π ∣ f ( x ) − g ( x ) ∣ d x ⋅ c m 2
Sur l’intervalle ] − π 2 , π 4 [ ]-\frac\pi2,\frac\pi4[ ] − 2 π , 4 π [ cos \cos cos sin \sin sin
Sur l’intervalle ] π 4 , 3 π 4 [ ]\frac\pi4,\frac{3\pi}4[ ] 4 π , 4 3 π [ sin \sin sin cos \cos cos
∫ − π 2 3 π 4 ∣ cos ( x ) − sin ( x ) ∣ d x = ∫ − π 2 π 4 ∣ cos ( x ) − sin ( x ) ∣ d x + ∫ π 4 3 π 4 ∣ cos ( x ) − sin ( x ) ∣ d x = ∫ − π 2 π 4 cos ( x ) − sin ( x ) d x − ∫ π 4 3 π 4 cos ( x ) − sin ( x ) d x = [ sin x + cos x ] − π 2 π 4 − [ sin x + cos x ] π 4 3 π 4 = 2 2 + 1 \begin{align*}
\int_{-\frac\pi2}^{\frac{3\pi}{4}}\left|\cos(x)-\sin(x)\right|dx
&=\int_{-\frac\pi2}^{\frac{\pi}{4}}\left|\cos(x)-\sin(x)\right|dx
+\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\left|\cos(x)-\sin(x)\right|dx \\
&=\int_{-\frac\pi2}^{\frac{\pi}{4}}\cos(x)-\sin(x)dx
-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\cos(x)-\sin(x)dx \\
&=\left[\sin x+\cos x\right]_{-\frac\pi2}^{\frac\pi4}
-\left[\sin x+\cos x\right]_{\frac\pi4}^{\frac{3\pi}4}
\\&=2\sqrt2+1
\end{align*} ∫ − 2 π 4 3 π ∣ cos ( x ) − sin ( x ) ∣ d x = ∫ − 2 π 4 π ∣ cos ( x ) − sin ( x ) ∣ d x + ∫ 4 π 4 3 π ∣ cos ( x ) − sin ( x ) ∣ d x = ∫ − 2 π 4 π cos ( x ) − sin ( x ) d x − ∫ 4 π 4 3 π cos ( x ) − sin ( x ) d x = [ sin x + cos x ] − 2 π 4 π − [ sin x + cos x ] 4 π 4 3 π = 2 2 + 1 A = ( 2 2 + 1 ) c m 2 \mathcal{A}=(2\sqrt2+1)cm^2 A = ( 2 2 + 1 ) c m 2