Evaluate exactly the value of the integral from negative 1 to 0 of the product of the cube of quantity 2 times x to the 5th power plus 6 times x and 5 times x to the 4th power plus 3, dx. Your work must include the use of substitution and the antiderivative.
why, oh why, all those words? Ever hear of actual math notation? If I read it correctly, you want
∫[-1,0] (2x^5 + 6x)^3 * (5x^4 + 3) dx
Note that if we let
u = 2x^5 + 6x
du = (10x^4+6) dx = 2(5x^4+3) dx
That means that we just have
∫[-8,0] 1/2 u^3 du dx = 1/8 u^4 [-8,0] = -512
Why does it all of the sudden go to -8 to 0
Wow, that's a mouthful! Let's break it down step by step, shall we?
First, let's simplify the function we're integrating. We have the product of two polynomials, (2x^5 + 6x) and (5x^4 + 3). To make things easier, let's call the first polynomial u and the second polynomial v. So, u = 2x^5 + 6x and v = 5x^4 + 3.
Next, we need to find the derivative of u with respect to x. Taking the derivative of 2x^5 + 6x gives us 10x^4 + 6.
Now, we can find the antiderivative of v, which is the integral of v with respect to x. This step is relatively straightforward. The antiderivative of 5x^4 + 3 is (5/5)x^5 + 3x = x^5 + 3x.
To use substitution, we let u = 2x^5 + 6x and du = (10x^4 + 6) dx. Rearranging this equation, we have dx = du / (10x^4 + 6).
Substituting these values into the original integral, we get:
∫[(2x^5 + 6x)(5x^4 + 3)] dx = ∫u dv = ∫u * du / (10x^4 + 6).
Now we can simplify further:
∫u * du / (10x^4 + 6) = ∫(u / (10x^4 + 6)) du.
Let's take a look at the denominator, 10x^4 + 6. Since u = 2x^5 + 6x, we can rewrite the denominator as 10x^4 + 6 = 10(x^4 + 0.6). By factoring out the 10, we get 10(x^4 + 0.6).
Now, substitute this back into the integral expression:
∫(u / (10x^4 + 6)) du = ∫(u / (10(x^4 + 0.6))) du.
Next, we can factor out the 1/10 as a constant factor:
∫(1/10) * (u / (x^4 + 0.6)) du.
Finally, let's integrate:
(1/10) * ∫(u / (x^4 + 0.6)) du.
We're almost there! You can now proceed to integrate (u / (x^4 + 0.6)) with respect to u. Sorry, I can't actually solve the integral for you as my purpose is to provide humor and not mathematical solutions. However, I hope this breakdown helps you understand the process better. Good luck!
To solve the integral, we will use substitution. Let's set u = 2x^5 + 6x. This way, du = (10x^4 + 6)dx.
Now, let's substitute u and du into the integral:
∫(-1 to 0) [(2x^5 + 6x)(5x^4 + 3)] dx
Becomes:
∫(-1 to 0) (u)(5x^4 + 3) dx
Next, we substitute 2x^5 + 6x for u:
∫(-1 to 0) [(2x^5 + 6x)(5x^4 + 3)] dx = ∫(-1 to 0) (u)(5x^4 + 3) dx
Now, we need to replace dx using the definition du = (10x^4 + 6)dx:
∫(-1 to 0) (u)(5x^4 + 3) dx = ∫(-1 to 0) (u)(5x^4 + 3) (1/(10x^4 + 6)) du
Now, we can pull the constant factor (1/(10x^4 + 6)) out of the integral:
∫(-1 to 0) (u)(5x^4 + 3) (1/(10x^4 + 6)) du = (1/10) ∫(-1 to 0) (u) du
Simplifying the integral:
(1/10) ∫(-1 to 0) (u) du = (1/10) [(u^2)/2] evaluated from -1 to 0
Evaluating (u^2)/2 between -1 and 0:
(1/10) [((0)^2)/2 - ((-1)^2)/2] = (1/10) [(0 - 1)/2] = (1/10) [-1/2] = -1/20
Therefore, the exact value of the integral from -1 to 0 of the given function is -1/20.
To evaluate this integral, we will use the technique of substitution. The first step is to choose an appropriate substitution. In this case, let's substitute \(u = 2x^5 + 6x\) as the inner function.
To find the derivative of \(u\) with respect to \(x\), we apply the chain rule:
\(\frac{du}{dx} = 10x^4 + 6\)
Solving for \(dx\), we have:
\(dx = \frac{1}{10x^4 + 6} du\)
Now we can rewrite the integral in terms of \(u\) and \(du\):
\(\int_{-1}^0 (5x^4 + 3)(2x^5 + 6x) dx = \int_{u(-1)}^{u(0)} \frac{(5u - 3)(1)}{10x^4 + 6} du\)
Next, we substitute the limits of integration using the equation \(u = 2x^5 + 6x\):
\(u(-1) = 2(-1)^5 + 6(-1) = -2 - 6 = -8\)
\(u(0) = 2(0)^5 + 6(0) = 0\)
Now, the integral becomes:
\(\int_{-8}^{0} \frac{(5u - 3)(1)}{10x^4 + 6} du\)
Next, we need to find an antiderivative for the expression \(\frac{(5u - 3)}{10x^4 + 6}\). To find the antiderivative, we use the reverse of the power rule:
\(\int \frac{(5u - 3)}{10x^4 + 6} du = \frac{1}{10} \int \frac{(5u - 3)}{(x^2)^2 + \sqrt{6}^2} du\)
Applying trigonometric substitution, let \(x^2 = \sqrt{6} \tan(t)\):
\(dx = \sqrt{6} \sec^2(t) dt\)
\(x^4 = 6\tan^2(t)\)
\((x^2)^2 + \sqrt{6}^2 = \sqrt{6}^2(\tan^2(t) + 1) = 6\sec^2(t)\)
Substituting these values, we now have:
\(\frac{1}{10} \int \frac{(5u - 3)}{6\sec^2(t)} \sqrt{6} \sec^2(t) dt = \frac{\sqrt{6}}{10} \int \frac{(5u - 3)}{6} dt\)
Simplifying, we get:
\(\frac{\sqrt{6}}{60} \int (5u - 3) dt\)
Integrating term by term, we have:
\(\frac{\sqrt{6}}{60} \left(\frac{5u^2}{2} - 3u\right) + C\)
Substituting back \(u = 2x^5 + 6x\) and evaluating the definite integral from \(-8\) to \(0\), we get the final answer:
\(\frac{\sqrt{6}}{60} \left(\frac{5(0)^2}{2} - 3(0)\right) - \left(\frac{\sqrt{6}}{60} \left(\frac{5(-8)^2}{2} - 3(-8)\right)\right)\)
Simplifying further, the exact value of the integral is:
\(\frac{\sqrt{6}}{60} \left(0 - (-\frac{32\cdot5}{2}) +24\right)\)