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How do you integrate the function x square e to the power x?
By using: ∫uv = u∫v - ∫u'∫v twice. ∫x2exdx let u = x2, v = ex, then: ∫x2exdx = x2∫exdx - ∫d/dx(x2)∫exdx dx = x2ex - ∫2xexdx Again, let u = 2x, v = ex, then: = x2ex - (2x∫exdx - ∫d/dx(2x)∫exdx dx) = x2ex - 2xex +∫2exdx = x2ex - 2xex + 2ex + C
What is x when its the subject of the equation c minus dx equals ex plus f?
ex+f = c -dx ex+dx = c -f x(e+d) = c -f x = c -f/(e+d)
What is the derivative of e raised to the 3x?
d/dx e3x = 3e3x
What does E stand for in E series?
e is a number equalling approximately 2.71828 It is special because it's derivative is the same as it [ie. d/dx (ex) = ex].
What is the natural logarithm of a variable x?
The natural logarithm of a variable x, is a variable y, such that ey = x. The constant e, is about 2.718281828, or more formally, e is a number such that the deriviative d/dx of ex = x.
What is the derivative of ex2?
It might become easier if you split the problem in half. Since we know that d/dx[f(x)-g(x)] is the same as the derivative of d/dx[f(x)] - d/dx[g(x)], you can rewrite your problem as d/dx(e)- d/dx(2x). (it's common usage to write coefficients in from of variables, but that's just notation). The derivative of e is simply 0 (since e is a constant) and the derivative of 2x is 2. so d/dx(e-2x)=0-2=-2
What is the Integral of 23x e2x dx?
This browser is pathetic for mathematical answers but here's the best that I can do:Let u = 23x therefore du/dx = 23let dv/dx = e^(2x) therefore v = 1/2*e^(2x)then, integrating by parts,I = I(u*dv/dx) dx = u*v - I(du/dx*v) dx= 23x*(1/2)*e^(2x) - I(23*(1/2)*e^(2x) dx= 23/2*xe^(2x) - 23/2*I(e^(2x)) dx= 23/2*xe^(2x) - 23/2*(1/2)*e^(2x)= 23/2*xe^(2x) - 23/4*e^(2x)or 23/4*e^(2x)*(2x - 1)
How do you integrate e power minus 2x?
∫e^(-2x) dx Let u = 2x du= 2 dx dx=(1/2) du ∫e^(-2x) dx = (1/2) ∫e^-u du = (1/2) (-e^-u) = -e^-u /2 + C = -e^-(2x) / 2 + C
How do you integrate the function x square e to the power x?
By using: ∫uv = u∫v - ∫u'∫v twice. ∫x2exdx let u = x2, v = ex, then: ∫x2exdx = x2∫exdx - ∫d/dx(x2)∫exdx dx = x2ex - ∫2xexdx Again, let u = 2x, v = ex, then: = x2ex - (2x∫exdx - ∫d/dx(2x)∫exdx dx) = x2ex - 2xex +∫2exdx = x2ex - 2xex + 2ex + C
How do you integrate of e -2x?
To integrate e^(-2x)dx, you need to take a u substitution. u=-2x du=-2dx Since the original integral does not have a -2 in it, you need to divide to get the dx alone. -(1/2)du=dx Since the integral of e^x is still e^x, you get: y = -(1/2)e^(-2x) Well, that was one method. I usually solve easier functions like this by thinking how the function looked like before it was differentiated. I let f(x) stand for the given function and F(x) stand for the primitive function; the function we had before differentiation (the integrated function). f(x)= e-2x <-- our given function F(x)= e-2x/-2 <-- our integrated function Evidence: F'(x)= -2e-2x/-2 = e-2x = f(x) Q.E.D It's as simple as that.
Solve the following differential equation:dy/dx = 3xy + x^3e^(x^2)?
To solve the differential equation dy/dx = 3xy + x^3e^(x^2), we can use the method of integrating factors. Here's the step-by-step solution: Step 1: Recognize the form of the differential equation as a first-order linear differential equation of the form dy/dx + P(x)y = Q(x), where P(x) = 3x and Q(x) = x^3e^(x^2). Step 2: Multiply the entire equation by the integrating factor, which is the exponential of the integral of P(x). In this case, the integrating factor is e^(∫3x dx) = e^(3x^2/2). Multiplying the equation by the integrating factor gives us: e^(3x^2/2) dy/dx + 3x e^(3x^2/2) y = x^3 e^(3x^2/2) * e^(x^2). Step 3: Notice that the left side of the equation can be simplified using the product rule for differentiation. The derivative of (e^(3x^2/2) y) with respect to x is given by d/dx (e^(3x^2/2) y) = e^(3x^2/2) dy/dx + 3x e^(3x^2/2) * y. Using this, the equation becomes: d/dx (e^(3x^2/2) y) = x^3 e^(4x^2/2). Step 4: Integrate both sides of the equation with respect to x: ∫d/dx (e^(3x^2/2) y) dx = ∫x^3 e^(4x^2/2) dx. This simplifies to: e^(3x^2/2) y = ∫x^3 e^(2x^2) dx. Step 5: Evaluate the integral on the right side of the equation: To solve the integral, we can use integration by parts. Let u = x^2 and dv = xe^(2x^2) dx. Then, du = 2x dx and v = (1/4) * e^(2x^2). Using integration by parts, the integral becomes: ∫x^3 e^(2x^2) dx = (1/4) x^2 e^(2x^2) - (1/4) ∫2x^2 * e^(2x^2) dx. Notice that we have another integral on the right side that is similar to the original integral. We can repeat the integration by parts process until we have an integral that we can solve. After integrating by parts twice, the integral becomes: ∫x^3 e^(2x^2) dx = (1/4) x^2 e^(2x^2) - (1/8) x e^(2x^2) + (1/16) ∫e^(2x^2) dx. The remaining integral, ∫e^(2x^2) dx, is a standard Gaussian integral and cannot be expressed in elementary functions. Therefore, we can denote it as √(π/8) * erf(x√2). Step 6: Substitute the evaluated integral back into the equation: e^(3x^2/2) y = (1/4) x^2 e^(2x^2) - (1/8) x * e^(2 x^2) + (1/16) √(π/8) erf(x√2). Step 7: Solve for y by dividing both sides of the equation by e^(3x^2/2): y = (1/4) x^2 e^(x^2) - (1/8) x e^(x^2) + (1/16) √(π/8) e^(-3x^2/2) * erf(x√2). Thus, the solution to the given differential equation is y = (1/4) x^2 e^(x^2) - (1/8) x e^(x^2) + (1/16) √(π/8) e^(-3x^2/2) * erf(x√2). Related More Question Visit : maths assignment help
What is the derivative of e power negative x?
d/dx (e-x) = -e-x
What is x when its the subject of the equation c minus dx equals ex plus f?
ex+f = c -dx ex+dx = c -f x(e+d) = c -f x = c -f/(e+d)
Can you solve the differential equation dy over dx equals y squared over e to the 2x power?
-2y square exp power -2x-1
What is the derivative of e raised to the 3x?
d/dx e3x = 3e3x
What is the lyre chords of on the floor by j-lo?
a,g,f,e,d,d,f,e,d,c,d,a,g,a(2x) its only chorus.....
What is the intergral of ex2?
Do you mean? int ex^2 dx This is a integration by parts. uv - int v du u = x^2 du = 2X dv = e v = e ex^2 - int e 2x ex^2 - ex^2 + C -------------------------
