Para encontrar el área bajo una curva en un intervalo [ a , b ] [a,\,b] [ a , b ]
A = ∫ a b f ( x ) d x A = \int_{a}^{b} f(x) \, dx A = ∫ a b f ( x ) d x
Encontrar el área bajo la curva f ( x ) = x + 5 f(x) = x + 5 f ( x ) = x + 5 [ 1 , 3 ] [1,\,3] [ 1 , 3 ]
Gráfica y tabulación
x x x y y y x + 5 x + 5 x + 5 0 0 0 5 5 5 0 + 5 0 + 5 0 + 5 1 1 1 6 6 6 1 + 5 1 + 5 1 + 5 2 2 2 7 7 7 2 + 5 2 + 5 2 + 5 3 3 3 8 8 8 3 + 5 3 + 5 3 + 5 4 4 4 9 9 9 4 + 5 4 + 5 4 + 5
Cálculo integral
∫ 1 3 ( x + 5 ) d x = x 2 2 + 5 x ∣ 1 3 = ( 3 ) 2 2 + 5 ( 3 ) − [ ( 1 ) 2 2 + 5 ( 1 ) ] = 9 2 + 15 − 1 2 − 5 = 8 2 + 10 = 14 \begin{aligned}
\int_{1}^{3} (x+5) \, dx &= \left. \frac{x^2}{2} + 5x \right|_{1}^{3} \\
&= \frac{(3)^2}{2} + 5(3) - \left[ \frac{(1)^2}{2} + 5(1) \right] \\
&= \frac{9}{2} + 15 - \frac{1}{2} - 5 \\
&= \frac{8}{2} + 10 \\
&= \mathbf{14}
\end{aligned} ∫ 1 3 ( x + 5 ) d x = 2 x 2 + 5 x 1 3 = 2 ( 3 ) 2 + 5 ( 3 ) − [ 2 ( 1 ) 2 + 5 ( 1 ) ] = 2 9 + 15 − 2 1 − 5 = 2 8 + 10 = 14
A = 14 u 2 A = 14\ u^2 A = 14 u 2
Encontrar el área bajo la curva en el intervalo indicado.
Bloque 1
f ( x ) = x + 4 f(x) = x + 4 f ( x ) = x + 4 en
[ 3 , 5 ] [3,\,5] [ 3 , 5 ] f ( x ) = x + 6 f(x) = x + 6 f ( x ) = x + 6 en
[ 2 , 4 ] [2,\,4] [ 2 , 4 ] f ( x ) = x + 2 f(x) = x + 2 f ( x ) = x + 2 en
[ 1 , 4 ] [1,\,4] [ 1 , 4 ]
Bloque 2
f ( x ) = − x + 3 f(x) = -x + 3 f ( x ) = − x + 3 en
[ 2 , 4 ] [2,\,4] [ 2 , 4 ] f ( x ) = − x + 5 f(x) = -x + 5 f ( x ) = − x + 5 en
[ 3 , 6 ] [3,\,6] [ 3 , 6 ] f ( x ) = − x + 10 f(x) = -x + 10 f ( x ) = − x + 10 en
[ 1 , 3 ] [1,\,3] [ 1 , 3 ]
Bloque 1
∫ 3 5 ( x + 4 ) d x = x 2 2 + 4 x ∣ 3 5 = ( 5 ) 2 2 + 4 ( 5 ) − [ ( 3 ) 2 2 + 4 ( 3 ) ] = 25 2 + 20 − 9 2 − 12 = 16 2 + 8 = 16 \begin{aligned}
\int_{3}^{5} (x+4) \, dx &= \left. \frac{x^2}{2} + 4x \right|_{3}^{5} \\
&= \frac{(5)^2}{2} + 4(5) - \left[ \frac{(3)^2}{2} + 4(3) \right] \\
&= \frac{25}{2} + 20 - \frac{9}{2} - 12 \\
&= \frac{16}{2} + 8 = \mathbf{16}
\end{aligned} ∫ 3 5 ( x + 4 ) d x = 2 x 2 + 4 x 3 5 = 2 ( 5 ) 2 + 4 ( 5 ) − [ 2 ( 3 ) 2 + 4 ( 3 ) ] = 2 25 + 20 − 2 9 − 12 = 2 16 + 8 = 16
∫ 2 4 ( x + 6 ) d x = x 2 2 + 6 x ∣ 2 4 = ( 4 ) 2 2 + 6 ( 4 ) − [ ( 2 ) 2 2 + 6 ( 2 ) ] = 16 2 + 24 − 4 2 − 12 = 12 2 + 12 = 18 \begin{aligned}
\int_{2}^{4} (x+6) \, dx &= \left. \frac{x^2}{2} + 6x \right|_{2}^{4} \\
&= \frac{(4)^2}{2} + 6(4) - \left[ \frac{(2)^2}{2} + 6(2) \right] \\
&= \frac{16}{2} + 24 - \frac{4}{2} - 12 \\
&= \frac{12}{2} + 12 = \mathbf{18}
\end{aligned} ∫ 2 4 ( x + 6 ) d x = 2 x 2 + 6 x 2 4 = 2 ( 4 ) 2 + 6 ( 4 ) − [ 2 ( 2 ) 2 + 6 ( 2 ) ] = 2 16 + 24 − 2 4 − 12 = 2 12 + 12 = 18
∫ 1 4 ( x + 2 ) d x = x 2 2 + 2 x ∣ 1 4 = ( 4 ) 2 2 + 2 ( 4 ) − [ ( 1 ) 2 2 + 2 ( 1 ) ] = 16 2 + 8 − 1 2 − 2 = 15 2 + 6 = 27 2 \begin{aligned}
\int_{1}^{4} (x+2) \, dx &= \left. \frac{x^2}{2} + 2x \right|_{1}^{4} \\
&= \frac{(4)^2}{2} + 2(4) - \left[ \frac{(1)^2}{2} + 2(1) \right] \\
&= \frac{16}{2} + 8 - \frac{1}{2} - 2 \\
&= \frac{15}{2} + 6 = \mathbf{\frac{27}{2}}
\end{aligned} ∫ 1 4 ( x + 2 ) d x = 2 x 2 + 2 x 1 4 = 2 ( 4 ) 2 + 2 ( 4 ) − [ 2 ( 1 ) 2 + 2 ( 1 ) ] = 2 16 + 8 − 2 1 − 2 = 2 15 + 6 = 2 27
Bloque 2
∫ 2 4 ( − x + 3 ) d x = − x 2 2 + 3 x ∣ 2 4 = − ( 4 ) 2 2 + 3 ( 4 ) − [ − ( 2 ) 2 2 + 3 ( 2 ) ] = − 16 2 + 12 + 4 2 − 6 = − 12 2 + 6 = 0 \begin{aligned}
\int_{2}^{4} (-x+3) \, dx &= \left. -\frac{x^2}{2} + 3x \right|_{2}^{4} \\
&= -\frac{(4)^2}{2} + 3(4) - \left[ -\frac{(2)^2}{2} + 3(2) \right] \\
&= -\frac{16}{2} + 12 + \frac{4}{2} - 6 \\
&= -\frac{12}{2} + 6 = \mathbf{0}
\end{aligned} ∫ 2 4 ( − x + 3 ) d x = − 2 x 2 + 3 x 2 4 = − 2 ( 4 ) 2 + 3 ( 4 ) − [ − 2 ( 2 ) 2 + 3 ( 2 ) ] = − 2 16 + 12 + 2 4 − 6 = − 2 12 + 6 = 0
∫ 3 6 ( − x + 5 ) d x = − x 2 2 + 5 x ∣ 3 6 = − ( 6 ) 2 2 + 5 ( 6 ) − [ − ( 3 ) 2 2 + 5 ( 3 ) ] = − 36 2 + 30 + 9 2 − 15 = − 27 2 + 15 = 3 2 \begin{aligned}
\int_{3}^{6} (-x+5) \, dx &= \left. -\frac{x^2}{2} + 5x \right|_{3}^{6} \\
&= -\frac{(6)^2}{2} + 5(6) - \left[ -\frac{(3)^2}{2} + 5(3) \right] \\
&= -\frac{36}{2} + 30 + \frac{9}{2} - 15 \\
&= -\frac{27}{2} + 15 = \mathbf{\frac{3}{2}}
\end{aligned} ∫ 3 6 ( − x + 5 ) d x = − 2 x 2 + 5 x 3 6 = − 2 ( 6 ) 2 + 5 ( 6 ) − [ − 2 ( 3 ) 2 + 5 ( 3 ) ] = − 2 36 + 30 + 2 9 − 15 = − 2 27 + 15 = 2 3
∫ 1 3 ( − x + 10 ) d x = − x 2 2 + 10 x ∣ 1 3 = − ( 3 ) 2 2 + 10 ( 3 ) − [ − ( 1 ) 2 2 + 10 ( 1 ) ] = − 9 2 + 30 + 1 2 − 10 = − 8 2 + 20 = 16 \begin{aligned}
\int_{1}^{3} (-x+10) \, dx &= \left. -\frac{x^2}{2} + 10x \right|_{1}^{3} \\
&= -\frac{(3)^2}{2} + 10(3) - \left[ -\frac{(1)^2}{2} + 10(1) \right] \\
&= -\frac{9}{2} + 30 + \frac{1}{2} - 10 \\
&= -\frac{8}{2} + 20 = \mathbf{16}
\end{aligned} ∫ 1 3 ( − x + 10 ) d x = − 2 x 2 + 10 x 1 3 = − 2 ( 3 ) 2 + 10 ( 3 ) − [ − 2 ( 1 ) 2 + 10 ( 1 ) ] = − 2 9 + 30 + 2 1 − 10 = − 2 8 + 20 = 16
Bloque A
f ( x ) = x + 3 f(x) = x + 3 f ( x ) = x + 3 en
[ 1 , 5 ] [1,\,5] [ 1 , 5 ] f ( x ) = x + 1 f(x) = x + 1 f ( x ) = x + 1 en
[ 2 , 6 ] [2,\,6] [ 2 , 6 ] f ( x ) = x + 7 f(x) = x + 7 f ( x ) = x + 7 en
[ 0 , 3 ] [0,\,3] [ 0 , 3 ]
Bloque B
f ( x ) = − x + 4 f(x) = -x + 4 f ( x ) = − x + 4 en
[ 1 , 3 ] [1,\,3] [ 1 , 3 ] f ( x ) = − x + 6 f(x) = -x + 6 f ( x ) = − x + 6 en
[ 2 , 5 ] [2,\,5] [ 2 , 5 ] f ( x ) = − x + 2 f(x) = -x + 2 f ( x ) = − x + 2 en
[ − 1 , 1 ] [-1,\,1] [ − 1 , 1 ]
Bloque C
f ( x ) = x + 8 f(x) = x + 8 f ( x ) = x + 8 en
[ 1 , 4 ] [1,\,4] [ 1 , 4 ] f ( x ) = x + 3 f(x) = x + 3 f ( x ) = x + 3 en
[ 2 , 5 ] [2,\,5] [ 2 , 5 ] f ( x ) = x + 5 f(x) = x + 5 f ( x ) = x + 5 en
[ 1 , 4 ] [1,\,4] [ 1 , 4 ]
Bloque D
f ( x ) = − x + 7 f(x) = -x + 7 f ( x ) = − x + 7 en
[ 3 , 5 ] [3,\,5] [ 3 , 5 ] f ( x ) = − x + 8 f(x) = -x + 8 f ( x ) = − x + 8 en
[ 2 , 6 ] [2,\,6] [ 2 , 6 ] f ( x ) = − x + 9 f(x) = -x + 9 f ( x ) = − x + 9 en
[ 4 , 7 ] [4,\,7] [ 4 , 7 ]