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Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
What is the antiderivative of -cscxcotx?
First, antiderivative = a solution to the indefinite integral therefore to integrate -(csc(x))(cot(x)) first convert it to -cos(x)/sin2(x) To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x) and du/dx = cosx This will make it ∫-1/u2 du and the antiderivative is 1/u +c, therefore the answer is 1/sin(x) + c.
Why is 10 to the power 1 equal to 10?
By definition of powers, x to the power y = x*x*x*........y times x to the power 1 = x Therefore, 10 to the power 1 = 10.
What is x to the negative 1 power times x to the negative 1 power?
x^-1 * x^-1 =x^(-1 + -1) =x^-2 or alternatively: 1/x times 1/x =1/x^2 (1/x^2) is the same as (x^-2)
How do you integrate open bracket x plus 1 close bracket cosx?
Try integration by parts, (twice, I think) with u=(x+1) and dv/dx=cosx
How do you integrate x power x?
e^x/1-e^x
How do you integrate x power e power x?
{xe^x dx integrate by parts let f(x) = x so f'(x) = 1 and g(x) = e^x so g'(x) = e^x so.. f(x)*g(x) - {(g(x)*f'(x)) dx therefore xe^x - {(e^x * 1) dx so.. xe^x - e^x + C factorize so... (x-1)e^x + C
Integrate square root X 4X from 1 to 2?
root x=x^(1/2) and 4x =4 x^1 you add the exponents then integrate as usual. The answer you should get is 4.
Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
How do you integrate 1 over x-1?
2
How do you integrate square root of 1 over x-1 dx?
2
What would be the integral of x where x has power 2?
If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.
Find general solution of x to power of 3 dp divided by dx equals a minus x?
I can't integrate a-x /x-3 ?
How you integrate xxxx 1 dx?
.2x^5+x+C
What is the antiderivative of e to the power of one divided by -x?
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
How do you find the derivative of a function that has been raised to a power greater than 1?
Let us say that f(x)=x^4A derivative is the opposite to an integral.If you were to integrate x^4, the first process is taking the power [which in this case is 4], multiplying it by any value before the x [which is 1], then subtracting 1 from the initial power [4]. This leaves 4x^3. The final step is taking the integral of what is 'inside' the power [which is (x)], and multiplying this to the entire answer, which results in 4x^3 x 1 = 4x^3If you were to derive (x)^4, you would just add 1 to the power [4] to become (x)^5 then put the value of the power as the denominator and the function as a numerator. This leaves [(x^5)/(5)]To assure that the derivative is correct, integrate it. (x^5) would become 5x^4. Since (x^5) is over (5), [(5x^4)/(5)] cancels the 5 on the numerator and denominator, thus leaving the original function of x^4
How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
