Answer on Question #61962 – Math – Calculus
Question
7 Determine J × 2 + 1 ( x + 2 ) 3 J \times 2 + 1(x + 2)3 J × 2 + 1 ( x + 2 ) 3
ln ( x + 2 ) + 4 x + 2 − 52 ( x + 3 ) 2 + c \ln(x + 2) + 4x + 2 - 52(x + 3)2 + c ln ( x + 2 ) + 4 x + 2 − 52 ( x + 3 ) 2 + c ln ( x + 2 ) − 4 x + 2 − 52 ( x + 3 ) 2 + c \ln(x + 2) - 4x + 2 - 52(x + 3)2 + c ln ( x + 2 ) − 4 x + 2 − 52 ( x + 3 ) 2 + c − ln ( x + 2 ) − 4 x + 2 − 52 ( x + 3 ) 2 + c - \ln(x + 2) - 4x + 2 - 52(x + 3)2 + c − ln ( x + 2 ) − 4 x + 2 − 52 ( x + 3 ) 2 + c ln ( x − 2 ) + 4 x − 2 − 52 ( x + 3 ) 2 + c \ln(x - 2) + 4x - 2 - 52(x + 3)2 + c ln ( x − 2 ) + 4 x − 2 − 52 ( x + 3 ) 2 + c Solution
Method 1
x 2 + 1 ( x + 2 ) 3 = A ( x + 2 ) 3 + B ( x + 2 ) 2 + C x + 2 , \frac{x^2 + 1}{(x + 2)^3} = \frac{A}{(x + 2)^3} + \frac{B}{(x + 2)^2} + \frac{C}{x + 2}, ( x + 2 ) 3 x 2 + 1 = ( x + 2 ) 3 A + ( x + 2 ) 2 B + x + 2 C ,
where
A = ( − 2 ) 2 + 1 = 4 + 1 = 5 , A = (-2)^2 + 1 = 4 + 1 = 5, A = ( − 2 ) 2 + 1 = 4 + 1 = 5 , B = ( x 2 + 1 ) ′ ∣ x = − 2 = ( 2 x ) ∣ x = − 2 = 2 ⋅ ( − 2 ) = − 4 , B = (x^2 + 1)'|_{x = -2} = (2x)|_{x = -2} = 2 \cdot (-2) = -4, B = ( x 2 + 1 ) ′ ∣ x = − 2 = ( 2 x ) ∣ x = − 2 = 2 ⋅ ( − 2 ) = − 4 , C = 1 2 ( x 2 + 1 ) ′ ′ ∣ x = − 2 = 1 2 ( 2 x ) ′ ∣ x = − 2 = 1 2 ⋅ 2 = 1. C = \frac{1}{2}(x^2 + 1)''|_{x = -2} = \frac{1}{2}(2x)'|_{x = -2} = \frac{1}{2} \cdot 2 = 1. C = 2 1 ( x 2 + 1 ) ′′ ∣ x = − 2 = 2 1 ( 2 x ) ′ ∣ x = − 2 = 2 1 ⋅ 2 = 1.
Thus, x 2 + 1 ( x + 2 ) 3 = A ( x + 2 ) 3 + B ( x + 2 ) 2 + C x + 2 = 5 ( x + 2 ) 3 − 4 ( x + 2 ) 2 + 1 x + 2 \frac{x^2 + 1}{(x + 2)^3} = \frac{A}{(x + 2)^3} + \frac{B}{(x + 2)^2} + \frac{C}{x + 2} = \frac{5}{(x + 2)^3} - \frac{4}{(x + 2)^2} + \frac{1}{x + 2} ( x + 2 ) 3 x 2 + 1 = ( x + 2 ) 3 A + ( x + 2 ) 2 B + x + 2 C = ( x + 2 ) 3 5 − ( x + 2 ) 2 4 + x + 2 1
Method 2
x 2 + 1 ( x + 2 ) 3 = A ( x + 2 ) 3 + B ( x + 2 ) 2 + C x + 2 = A + B ( x + 2 ) + C ( x + 2 ) 2 ( x + 2 ) 3 , \frac{x^2 + 1}{(x + 2)^3} = \frac{A}{(x + 2)^3} + \frac{B}{(x + 2)^2} + \frac{C}{x + 2} = \frac{A + B(x + 2) + C(x + 2)^2}{(x + 2)^3}, ( x + 2 ) 3 x 2 + 1 = ( x + 2 ) 3 A + ( x + 2 ) 2 B + x + 2 C = ( x + 2 ) 3 A + B ( x + 2 ) + C ( x + 2 ) 2 ,
hence A + B ( x + 2 ) + C ( x + 2 ) 2 = x 2 + 1 A + B(x + 2) + C(x + 2)^2 = x^2 + 1 A + B ( x + 2 ) + C ( x + 2 ) 2 = x 2 + 1
If x = − 2 x = -2 x = − 2 A = ( − 2 ) 2 + 1 = 4 + 1 = 5 A = (-2)^2 + 1 = 4 + 1 = 5 A = ( − 2 ) 2 + 1 = 4 + 1 = 5
5 + B ( x + 2 ) + C ( x + 2 ) 2 = x 2 + 1 , hence 5 + B(x + 2) + C(x + 2)^2 = x^2 + 1, \text{ hence} 5 + B ( x + 2 ) + C ( x + 2 ) 2 = x 2 + 1 , hence B ( x + 2 ) + C ( x + 2 ) 2 = x 2 + 1 − 5 , B(x + 2) + C(x + 2)^2 = x^2 + 1 - 5, B ( x + 2 ) + C ( x + 2 ) 2 = x 2 + 1 − 5 , B ( x + 2 ) + C ( x + 2 ) 2 = x 2 − 4. B(x + 2) + C(x + 2)^2 = x^2 - 4. B ( x + 2 ) + C ( x + 2 ) 2 = x 2 − 4.
If x = − 1 x = -1 x = − 1 B + C = ( − 1 ) 2 − 4 = 1 − 4 = − 3 B + C = (-1)^2 - 4 = 1 - 4 = -3 B + C = ( − 1 ) 2 − 4 = 1 − 4 = − 3
If x = 0 x = 0 x = 0 2 B + 4 C = 0 2 − 4 = 0 − 4 = − 4 2B + 4C = 0^2 - 4 = 0 - 4 = -4 2 B + 4 C = 0 2 − 4 = 0 − 4 = − 4
Solving the system
{ B + C = − 3 , 2 B + 4 C = − 4 , \left\{
\begin{array}{l}
B + C = -3, \\
2B + 4C = -4,
\end{array}
\right. { B + C = − 3 , 2 B + 4 C = − 4 ,
Dividing the second equation by 2
{ B + C = − 3 , B + 2 C = − 2 , \left\{ \begin{array}{l} B + C = - 3, \\ B + 2 C = - 2, \end{array} \right. { B + C = − 3 , B + 2 C = − 2 ,
Subtracting the first equation from the second one
{ B + C = − 3 , B + 2 C − ( B + C ) = − 2 − ( − 3 ) , \left\{ \begin{array}{c} B + C = - 3, \\ B + 2 C - (B + C) = - 2 - (- 3), \end{array} \right. { B + C = − 3 , B + 2 C − ( B + C ) = − 2 − ( − 3 ) , { B + C = − 3 , C = 1 , \left\{ \begin{array}{c} B + C = - 3, \\ C = 1, \end{array} \right. { B + C = − 3 , C = 1 ,
Plugging C = 1 C = 1 C = 1
{ B + 1 = − 3 , C = 1 , \left\{ \begin{array}{l} B + 1 = - 3, \\ C = 1, \end{array} \right. { B + 1 = − 3 , C = 1 , { B = − 3 − 1 , C = 1 , \left\{ \begin{array}{c} B = - 3 - 1, \\ C = 1, \end{array} \right. { B = − 3 − 1 , C = 1 , { B = − 4 , C = 1 , \left\{ \begin{array}{l} B = - 4, \\ C = 1, \end{array} \right. { B = − 4 , C = 1 ,
Thus, x 2 + 1 ( x + 2 ) 3 = A ( x + 2 ) 3 + B ( x + 2 ) 2 + C x + 2 = 5 ( x + 2 ) 3 − 4 ( x + 2 ) 2 + 1 x + 2 \frac{x^2 + 1}{(x + 2)^3} = \frac{A}{(x + 2)^3} + \frac{B}{(x + 2)^2} + \frac{C}{x + 2} = \frac{5}{(x + 2)^3} - \frac{4}{(x + 2)^2} + \frac{1}{x + 2} ( x + 2 ) 3 x 2 + 1 = ( x + 2 ) 3 A + ( x + 2 ) 2 B + x + 2 C = ( x + 2 ) 3 5 − ( x + 2 ) 2 4 + x + 2 1
**Method 3**
x 2 + 1 ( x + 2 ) 3 = A ( x + 2 ) 3 + B ( x + 2 ) 2 + C x + 2 \frac {x ^ {2} + 1}{(x + 2) ^ {3}} = \frac {A}{(x + 2) ^ {3}} + \frac {B}{(x + 2) ^ {2}} + \frac {C}{x + 2} ( x + 2 ) 3 x 2 + 1 = ( x + 2 ) 3 A + ( x + 2 ) 2 B + x + 2 C x 2 + 1 = x 2 + 4 x − 4 x + 4 − 4 − 8 + 8 + 1 = x 2 + 4 x + 4 − 4 x − 8 − 4 + 8 + 1 = = ( x 2 + 4 x + 4 ) − 4 ( x + 2 ) + ( 8 − 4 + 1 ) = = ( x + 2 ) 2 − 4 ( x + 2 ) + 5 \begin{array}{l} x ^ {2} + 1 = x ^ {2} + 4 x - 4 x + 4 - 4 - 8 + 8 + 1 = x ^ {2} + 4 x + 4 - 4 x - 8 - 4 + 8 + 1 = \\ = (x ^ {2} + 4 x + 4) - 4 (x + 2) + (8 - 4 + 1) = \\ = (x + 2) ^ {2} - 4 (x + 2) + 5 \\ \end{array} x 2 + 1 = x 2 + 4 x − 4 x + 4 − 4 − 8 + 8 + 1 = x 2 + 4 x + 4 − 4 x − 8 − 4 + 8 + 1 = = ( x 2 + 4 x + 4 ) − 4 ( x + 2 ) + ( 8 − 4 + 1 ) = = ( x + 2 ) 2 − 4 ( x + 2 ) + 5
Using synthetic division
Hence
x 2 + 1 ( x + 2 ) 3 = ( x + 2 ) 2 − 4 ( x + 2 ) + 5 ( x + 2 ) 3 = ( x + 2 ) 2 ( x + 2 ) 3 − 4 x + 2 ( x + 2 ) 3 + 5 1 ( x + 2 ) 3 = = 1 x + 2 − 4 ⋅ 1 ( x + 2 ) 2 + 5 ( x + 2 ) 3 . \begin{array}{l} \frac {x ^ {2} + 1}{(x + 2) ^ {3}} = \frac {(x + 2) ^ {2} - 4 (x + 2) + 5}{(x + 2) ^ {3}} = \frac {(x + 2) ^ {2}}{(x + 2) ^ {3}} - 4 \frac {x + 2}{(x + 2) ^ {3}} + 5 \frac {1}{(x + 2) ^ {3}} = \\ = \frac {1}{x + 2} - 4 \cdot \frac {1}{(x + 2) ^ {2}} + \frac {5}{(x + 2) ^ {3}}. \\ \end{array} ( x + 2 ) 3 x 2 + 1 = ( x + 2 ) 3 ( x + 2 ) 2 − 4 ( x + 2 ) + 5 = ( x + 2 ) 3 ( x + 2 ) 2 − 4 ( x + 2 ) 3 x + 2 + 5 ( x + 2 ) 3 1 = = x + 2 1 − 4 ⋅ ( x + 2 ) 2 1 + ( x + 2 ) 3 5 .
Finally obtain
∫ x 2 + 1 ( x + 2 ) 3 d x = ∫ ( 1 x + 2 − 4 ( x + 2 ) 2 + 5 ( x + 2 ) 3 ) d ( x + 2 ) = ln ( x + 2 ) + 4 x + 2 − 5 2 ( x + 2 ) 2 + c , \int \frac {x ^ {2} + 1}{(x + 2) ^ {3}} d x = \int \left(\frac {1}{x + 2} - \frac {4}{(x + 2) ^ {2}} + \frac {5}{(x + 2) ^ {3}}\right) d (x + 2) = \ln (x + 2) + \frac {4}{x + 2} - \frac {5}{2 (x + 2) ^ {2}} + c, ∫ ( x + 2 ) 3 x 2 + 1 d x = ∫ ( x + 2 1 − ( x + 2 ) 2 4 + ( x + 2 ) 3 5 ) d ( x + 2 ) = ln ( x + 2 ) + x + 2 4 − 2 ( x + 2 ) 2 5 + c ,
where c c c
Answer: ln ( x + 2 ) + 4 x + 2 − 5 2 ( x + 2 ) 2 + c . \ln(x + 2) + \frac{4}{x + 2} - \frac{5}{2(x + 2)^2} + c. ln ( x + 2 ) + x + 2 4 − 2 ( x + 2 ) 2 5 + c .
Question
8 Find the volume of a sphere generated by a semicircle y = ( r 2 − x 2 ) y = (\sqrt{r^2 - x^2}) y = ( r 2 − x 2 )
- π \pi π
- 4 π r 32 4\pi r32 4 π r 32
- π r 34 \pi r34 π r 34
- 4 π r 33 4\pi r33 4 π r 33
Solution
V = ∫ − r r π y 2 d x = ∫ − r r π ( r 2 − x 2 ) d x = π ( r 2 x − x 3 3 ) − r r = π ( 2 3 r 3 ) − π ( − 2 3 r 3 ) = 4 3 π r 3 . V = \int_{-r}^{r} \pi y^2 dx = \int_{-r}^{r} \pi (r^2 - x^2) dx = \pi \left(r^2 x - \frac{x^3}{3}\right)_{-r}^{r} = \pi \left(\frac{2}{3} r^3\right) - \pi \left(-\frac{2}{3} r^3\right) = \frac{4}{3} \pi r^3. V = ∫ − r r π y 2 d x = ∫ − r r π ( r 2 − x 2 ) d x = π ( r 2 x − 3 x 3 ) − r r = π ( 3 2 r 3 ) − π ( − 3 2 r 3 ) = 3 4 π r 3 .
Answer: 4 3 π r 3 . \frac{4}{3} \pi r^3. 3 4 π r 3 .
Question
9 Evaluate ∫ x 2 e 3 x \int x^2 e^3 x ∫ x 2 e 3 x
- e 3 x 3 ( x 2 − 2 x 3 + 29 ) + c e^3 x^3(x^2 - 2x^3 + 29) + c e 3 x 3 ( x 2 − 2 x 3 + 29 ) + c
- − e 3 x 3 ( x 2 + 2 x 3 − 29 ) + c -e^3 x^3(x^2 + 2x^3 - 29) + c − e 3 x 3 ( x 2 + 2 x 3 − 29 ) + c
- e 2 x 3 ( x 3 − x 4 + 29 ) + c e^2 x^3(x^3 - x^4 + 29) + c e 2 x 3 ( x 3 − x 4 + 29 ) + c
- e 3 x 2 + 2 x 3 − 25 ) + c e^3 x^2 + 2x^3 - 25) + c e 3 x 2 + 2 x 3 − 25 ) + c
Solution
Using integration by parts
∫ x 2 e 3 x d x = 1 3 x 2 e 3 x − ∫ 2 3 x e 3 x d x = 1 3 x 2 e 3 x − 2 3 ∫ x e 3 x d x = 1 3 x 2 e 3 x − 2 3 [ 1 3 x e 3 x − ∫ 1 3 e 3 x d x ] = = 1 3 x 2 e 3 x − 2 3 [ 1 3 x e 3 x − 1 9 e 3 x ] + c = 1 3 x 2 e 3 x − 2 9 x e 3 x + 2 27 e 3 x + c = 9 27 x 2 e 3 x − 6 27 x e 3 x + 2 27 e 3 x + c = = 1 27 ( 9 x 2 − 6 x + 2 ) e 3 x + c , \begin{aligned}
\int x^2 e^{3x} dx &= \frac{1}{3} x^2 e^{3x} - \int \frac{2}{3} x e^{3x} dx = \frac{1}{3} x^2 e^{3x} - \frac{2}{3} \int x e^{3x} dx = \frac{1}{3} x^2 e^{3x} - \frac{2}{3} \left[ \frac{1}{3} x e^{3x} - \int \frac{1}{3} e^{3x} dx \right] = \\
&= \frac{1}{3} x^2 e^{3x} - \frac{2}{3} \left[ \frac{1}{3} x e^{3x} - \frac{1}{9} e^{3x} \right] + c = \frac{1}{3} x^2 e^{3x} - \frac{2}{9} x e^{3x} + \frac{2}{27} e^{3x} + c = \frac{9}{27} x^2 e^{3x} - \frac{6}{27} x e^{3x} + \frac{2}{27} e^{3x} + c = \\
&= \frac{1}{27} (9x^2 - 6x + 2) e^{3x} + c,
\end{aligned} ∫ x 2 e 3 x d x = 3 1 x 2 e 3 x − ∫ 3 2 x e 3 x d x = 3 1 x 2 e 3 x − 3 2 ∫ x e 3 x d x = 3 1 x 2 e 3 x − 3 2 [ 3 1 x e 3 x − ∫ 3 1 e 3 x d x ] = = 3 1 x 2 e 3 x − 3 2 [ 3 1 x e 3 x − 9 1 e 3 x ] + c = 3 1 x 2 e 3 x − 9 2 x e 3 x + 27 2 e 3 x + c = 27 9 x 2 e 3 x − 27 6 x e 3 x + 27 2 e 3 x + c = = 27 1 ( 9 x 2 − 6 x + 2 ) e 3 x + c ,
where c c c
Answer: 1 27 ( 9 x 2 − 6 x + 2 ) e 3 x + c \frac{1}{27} (9x^2 - 6x + 2) e^{3x} + c 27 1 ( 9 x 2 − 6 x + 2 ) e 3 x + c
Question
10 Given f ( x ) = ( 7 x 4 − 5 x 3 ) f(x) = (7x^4 - 5x^3) f ( x ) = ( 7 x 4 − 5 x 3 ) d f ( x ) d x \frac{df(x)}{dx} d x df ( x )
- 7 x 4 − 5 x 3 7x^4 - 5x^3 7 x 4 − 5 x 3
- 2 x 3 − 15 x 2 2x^3 - 15x^2 2 x 3 − 15 x 2
28x2-15x2
28x3-15x2
Solution
d f ( x ) d x = d d x ( 7 x 4 − 5 x 3 ) = 7 d d x ( x 4 ) − 5 d d x ( x 3 ) = 7 ⋅ ( 4 x 3 ) − 5 ⋅ ( 3 x 2 ) = 28 x 3 − 15 x 2 . \frac {\mathrm {d f} (\mathrm {x})}{\mathrm {d x}} = \frac {d}{d x} (7 \mathrm {x} ^ {4} - 5 \mathrm {x} ^ {3}) = 7 \frac {d}{d x} (\mathrm {x} ^ {4}) - 5 \frac {d}{d x} (\mathrm {x} ^ {3}) = 7 \cdot (4 x ^ {3}) - 5 \cdot (3 x ^ {2}) = 2 8 x ^ {3} - 1 5 x ^ {2}. dx df ( x ) = d x d ( 7 x 4 − 5 x 3 ) = 7 d x d ( x 4 ) − 5 d x d ( x 3 ) = 7 ⋅ ( 4 x 3 ) − 5 ⋅ ( 3 x 2 ) = 28 x 3 − 15 x 2 .
Answer: 28 x 3 − 15 x 2 28x^{3} - 15x^{2} 28 x 3 − 15 x 2
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#340153 on Dec 2023