Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(x) + C
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
The integral of cot(x)dx is ln|sin(x)| + C
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First see if the integrand ie x2-4 is negative anywhere in the range (it might change sign either side of anyplace where the function is zero, so first solve for x2-4=0). If it is, reverse the sign in that part. Here if x<2 it is negative so the modulus gives 4-x2. So integrate 4-x2 from 0 to 2 and then integrate x2-4 from 2 to 3. Add the 2 results together.
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.
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x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
Yes. Let us use an example: x=1:10; % Data Set z=1:20; % Range y=x^2; % Equation equation1=trapz(y,z) % This will integrate y over the range of z for your data When integrating over a matrix: x2=[ 1 2 3; 4 5 6]; % Matrix y2=x*C; % Where c is any variable equation2=trapz(y2); % Will integrate over every column of your matrix and give you an array.
You will have to use partial fractions for this one. Split up the fraction into two simpler fractions, of the form A / x + B / (4-x). The result will be easy to integrate.
∫[√(4x) / x] dx = ∫(2 / √x)dx = 2∫(x-1/2) dx = 2(2x1/2 + C) = 4√x + C
e^x/1-e^x
4 is a constant, so you can pull it out of the integral. Use a u substitution with u = 3-x and du = -dx. If it's a definite integral, remember to change the limits of integration. The integral is then simply 1/u which integrates to be ln u. Substitute back in 3-x for u and you have the answer to be: -4*ln (3-x) + C
-(10/x)
x/4 - 4 = x/4 4x/4 - 4 x 4 = x x - 16 = x 0 = 16 There is no solution to the equation.
Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(x) + C