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In order to work out this problem, we need to learn how to apply the integration method correctly.
The given expression is ∫ 2xln(2x) dx.
Instead of working out with 2x's, we let u = 2x. Then, du = 2 dx or du/2 = dx. This method is both valid and easy to avoid working out with too much expressions. You should get:
∫ uln(u) (du/2)
= ½ ∫ uln(u) du
Use integration by parts, which states that:
∫ f(dg) = fg - ∫ g(df)
We let:
- f = ln(u). Then, df = 1/u du
- dg = u du. Then, g = ∫ u du = ½u²
Using these substitutions, we now have:
½(½u²ln(u) - ½∫ u du)
= ¼(u²ln(u) - ∫ u du)
Finally, by integration, we obtain:
¼ * (u²ln(u) - ½u²) + c
= 1/8 * (2u²ln(u) - u²) + c
= 1/8 * (2(2x)²ln(u) - (2x)²) + c
= 1/8 * (2x)² * (2ln(u) - 1) + c
= ½ * x² * (2ln(2x - 1)) + c
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What is the indefinite integral?
An indefinite integral is a version of an integral that, unlike a definite integral, returns an expression instead of a number. The general form of a definite integral is: ∫ba f(x) dx. The general form of an indefinite integral is: ∫ f(x) dx. An example of a definite integral is: ∫20 x2 dx. An example of an indefinite integral is: ∫ x2 dx In the definite case, the answer is 23/3 - 03/3 = 8/3. In the indefinite case, the answer is x3/3 + C, where C is an arbitrary constant.
What is the integral of 2-2x with limits 0 to t?
Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->t) (2-2x) dy would be different, you must state integrate with respect to what otherwise it can be anything) So integrals preserves sum and product with constants. i.e. Integrate (2-2x) dx = Integrate 2 dx - 2integrate x dx = 2x - x^2 + C By Fundamental Theorem of Calculus, take any anti-derivative, say C = 0 would be fine, and Integral(0->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
How do you evaluate the integral of 2udu divided by 1 plus u squared?
for the integral of (2x)dx/(1+x2 ) Take (1+x2 ) as your 'u' substitution. find du, du= 2x dx use u substitution to write new integral, integral of du/u the integral of du/u= ln abs(u) + C therefore, your original problem becomes an answer with ln ln abs (1+x2) + C *abs refers to absolute value of the parentheses
What is the integral of 2?
The integral of 2 is "who gives a $%&#." You need to know what 2 is relating to and what the 2 means. If the question is "What is the integral of 2 dx" the answer would be "2x + c," with c being a constant. If you instead wish to know what the integral of 2 dy, the answer is very different. (2y +c)
How do you graph a tangent function y equals 2tan 2x?
The derivative of a function is the function's tangent function.So, d(y)/dx = d(2tan(2x))/dx = 2*d(tan(2x))/dx = 2*d(tan(u))/du*du/dx, where u=2x, = 2*sec2(2x)*d(2x)/dx = 4*sec2(2x)*d(x)/dx = 4*sec2(2x)Now just make a plot for y = 4*sec2(2x) and you got your tangent function.
