How to evaluate the integral

Mathematics

1 answer:

Papessa [141]3 years ago

8 0

Remember
for F(x) is the antiderivitive of f(x)
$\int\limits^a_b {f(x)} \, dx =F(a)-F(b)$
so find the antiderivitive of ((x+1)^2)/x

if we expand we get (x^2+2x+1)/x which simplifies to x+2+(1/x)
the anti-deritivive of x is (1/2)x^2
the antideritiveve of 2 is 2x
the antideritivieve of 1/x is ln|x|

F(x)=(1/2)x^2+2x+ln|x|+C

$\int\limits^2_1 { \frac{(x+1)^2}{x} } \, dx =F(2)-F(1)$

F(1)=(5/2)+ln1+C
F(2)=6+ln2+C

F(2)-F(1)=6+ln2+C-(5/2+ln1+C)
F(2)-F(1)=(7/2)+ln2
that is the answer
if you want is simplified or expanded it is about 4.1915

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X-2/x+4=x-6/x-4 how to solve ? a. x=8 b. x=3 c. x=9 d. x=5

lana66690 [7]

Answer:

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3 years ago

Determine the formula for the nth term of the sequence: -2,1,7,25,79,...

rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

$x_n = a n^4 + b n^3 + c n^2 + d n + e$

From the given known values of the sequence, we have

$\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}$

Solving the system yields coefficients

$a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4$

so that the n-th term in the sequence might be

$\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}$

Then the next few terms in the sequence could very well be

$\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}$

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of $\{x_n\}$ eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of $\{x_n\}$ by $\Delta^{k}\{x_n\}$ . Then

• 1st-order differences:

$\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}$

• 2nd-order differences:

$\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}$

• 3rd-order differences:

$\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}$

• 4th-order differences:

$\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}$

From here I made the assumption that $\Delta^4\{x_n\}$ is the constant sequence {15, 15, 15, …}. This implies $\Delta^3\{x_n\}$ forms an arithmetic/linear sequence, which implies $\Delta^2\{x_n\}$ forms a quadratic sequence, and so on up $\{x_n\}$ forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.