Evaluate the function below from x = 1 up to x = 4.
∫(7−lnx(x)(3+lnx))dx
Suspicious of your typing, especially the lnx(x) part
Sent it to Wolfram and they read it as:
https://www.wolframalpha.com/input/?i=%E2%88%AB%287%E2%88%92lnx%28x%29%283%2Blnx%29%29dx+
https://www.wolframalpha.com/input/?i=%E2%88%AB%287%E2%88%92lnx%28x%29%283%2Blnx%29%29dx+%2C+from+1+to+4
Wolfram uses log for ln
To integrate lnx or x lnx, use integration by parts
u = lnx, du = 1/x dx
dv = dx, v = x
∫lnx dx = x lnx - ∫1/x * x dx = x lnx - x
If
u = lnx, du = 1/x dx
dv = x dx, v = 1/2 x^2
∫x lnx = 1/2 x^2 lnx - ∫1/2 x dx = 1/2 x^2 lnx - 1/4 x^2
Well, well, well, are you ready for some integration excitement? Let's dive into it!
To evaluate this integral ∫(7−lnx(x)(3+lnx))dx from x = 1 up to x = 4, we need to find the antiderivative of the function first. It's like hunting for clues to unlock the secret treasure of integration!
Grab your binoculars, mate, because we've got logarithms and multiplication malarkey going on here. Let's break this down step by step.
First, distribute the x and simplify the expressions:
∫(7 - xlnx(3 + lnx)) dx
Next, integrate term by term. This part requires some cleverness:
∫7 dx = 7x
∫xlnx dx: Hmm, this one is tricky. Let me think for a moment... Ah! I know! We can use integration by parts:
u = ln x (because it's easy to differentiate)
dv = x dx (because it's easy to integrate)
du = (1/x) dx (the derivative of ln x)
v = (1/2)x^2 (the antiderivative of x)
Using the formula for integration by parts:
∫u dv = uv - ∫v du
We can substitute in for u and v:
∫xlnx dx = (1/2)x^2 ln x - ∫(1/2)x^2 (1/x) dx
Simplifying the second integral:
∫xlnx dx = (1/2)x^2 ln x - (1/2)∫x dx
= (1/2)x^2 ln x - (1/2)(1/2)x^2 + C
∫(1/2)x^2 ln x dx = (1/2)x^2 ln x - (1/4)x^2
Lastly, let's integrate the last term:
∫ -xlnx^2 dx: Looks like another one for integration by parts!
u = ln x^2 (because it's easy to differentiate)
dv = -x dx (because it's easy to integrate)
du = 2(1/x) dx (the derivative of ln x^2)
v = -(1/2)x^2 (the antiderivative of -x)
Using the formula for integration by parts:
∫u dv = uv - ∫v du
We can substitute in for u and v:
∫-xlnx^2 dx = -(1/2)x^2 ln x^2 - ∫-(1/2)x^2 (2/x) dx
Simplifying the integral:
∫-xlnx^2 dx = -(1/2)x^2 ln x^2 + ∫x dx
= -(1/2)x^2 ln x^2 + (1/2)x^2 + C
Now that we have all the pieces, let's put them together:
∫(7−lnx(x)(3+lnx))dx = 7x - (1/2)x^2 ln x - (1/4)x^2 - -(1/2)x^2 ln x^2 + (1/2)x^2 + C
= 7x - (1/2)x^2 ln x - (1/4)x^2 - (1/2)x^2 ln x^2 + (1/2)x^2 + C
= 7x - (1/2)x^2 ln x - (1/4)x^2 - (1/2)x^2 ln x^2 + (1/2)x^2 + C
= 7x - (1/2)x^2 ln x - (1/2)x^2 ln x^2 + (1/2)x^2 - (1/4)x^2 + C
= 7x - (1/2)x^2 ln x - (1/2)x^2 ln x^2 + (1/4)x^2 + C
Now, we can evaluate the definite integral by plugging in the limits of integration:
∫(7−lnx(x)(3+lnx))dx from x = 1 up to x = 4:
[7(4) - (1/2)(4)^2 ln(4) - (1/2)(4)^2 ln(4)^2 + (1/4)(4)^2] - [7(1) - (1/2)(1)^2 ln(1) - (1/2)(1)^2 ln(1)^2 + (1/4)(1)^2]
Simplifying calculations, savoring each digit:
[28 - 8 ln(4) - 8 ln(16) + 4] - [7 - 0 - 0 + 1]
28 - 8 ln(4) - 8 ln(16) + 4 - 7 + 1
(28 + 4 - 7 + 1) - 8 ln(4) - 8 ln(16)
26 - 8 ln(4) - 8 ln(16)
Congratulations! You've reached the final answer of the integral: 26 - 8 ln(4) - 8 ln(16).
To evaluate the function ∫(7−lnx(x)(3+lnx))dx from x = 1 up to x = 4, we need to find the definite integral of the function over this range.
1. To find the antiderivative of the given function, let's split it into separate terms. We have:
∫(7−lnx(x)(3+lnx))dx = ∫(7 - 3lnx - ln^2x) dx
2. Take the integral of each term separately. The integral of 7 with respect to x is 7x. The integral of 3lnx with respect to x is 3xlnx - 3x. For the integral of ln^2x, we can rewrite it as (lnx)^2 and use integration by parts to solve it.
Letting u = lnx and dv = lnx dx,
we get du = 1/x dx and v = xlnx - x.
By integration by parts: ∫(lnx)^2 dx = xlnx - x - ∫x(1/x)dx = xlnx - x + C,
where C is the constant of integration.
Combining all the terms, we have:
∫(7−lnx(x)(3+lnx))dx = 7x - 3xlnx + 3x - xlnx + x - C.
3. Now we can evaluate the definite integral from x = 1 to x = 4. Substituting x = 4 and x = 1 into the antiderivative expression, we get:
[7(4) - 3(4)ln(4) + 3(4) - 4ln(4) + 4 - C] - [7(1) - 3(1)ln(1) + 3(1) - 1ln(1) + 1 - C]
Simplifying, we have:
[28 - 12ln(4) + 12 - 4ln(4) + 4 - C] - [7 - C]
= 28 - 12ln(4) + 12 - 4ln(4) + 4 - C - 7 + C
= 32 - 16ln(4)
Therefore, the value of the given integral from x = 1 up to x = 4 is 32 - 16ln(4).
To evaluate the given function ∫(7−lnx(x)(3+lnx))dx from x = 1 to x = 4, we need to perform definite integration.
First, let's simplify the function before integrating. We have:
∫(7−lnx(x)(3+lnx))dx
Expanding the expression, we get:
∫(7x − 3ln(x)x − ln^2(x)x)dx
Simplifying further, we have:
∫(7x^2 - 3xln(x) - ln^2(x)x)dx
Now, we can integrate the simplified expression. Let's break it down into three separate integrals:
∫(7x^2)dx - ∫(3xln(x))dx - ∫(ln^2(x)x)dx
Step 1: Evaluating the first integral
∫(7x^2)dx = (7/3)x^3
Step 2: Evaluating the second integral
∫(3xln(x))dx = (3/2)x^2ln(x) - (3/2)∫(x)dx
= (3/2)x^2ln(x) - (3/2)(1/2)x^2 + C1
= (3/2)x^2ln(x) - (3/4)x^2 + C1
Step 3: Evaluating the third integral
∫(ln^2(x)x)dx = (1/3)x^3ln^2(x) - (2/3)∫(x^2ln(x))dx
To evaluate ∫(x^2ln(x))dx, we can use integration by parts. Let's assume u = ln(x) and dv = x^2 dx.
du = (1/x) dx and v = (1/3)x^3
Using the integration by parts formula:
∫(x^2ln(x))dx = uv - ∫vdu
= (1/3)x^3ln(x) - ∫((1/3)x^2)dx
= (1/3)x^3ln(x) - (1/9)x^3 + C2
Note: C1 and C2 represent the constants of integration.
Combining all the results, the integral of the given function ∫(7−lnx(x)(3+lnx))dx from x = 1 to x = 4 is:
[(7/3)x^3 - (3/2)x^2ln(x) + (3/4)x^2 - (1/3)x^3ln^2(x) + (1/9)x^3] evaluated from x = 1 to x = 4.
Substituting the values of x:
[(7/3)(4)^3 - (3/2)(4)^2ln(4) + (3/4)(4)^2 - (1/3)(4)^3ln^2(4) + (1/9)(4)^3] - [(7/3)(1)^3 - (3/2)(1)^2ln(1) + (3/4)(1)^2 - (1/3)(1)^3ln^2(1) + (1/9)(1)^3]
After evaluating this expression, you will get the final result of the definite integral.