Find dy/dx and d2y/dx2 if y= definite integral sign where a= 1 and b= 3x
1/(t^2+t+1) dt
how do i even start. do i integrate and then plug in a and b? plz help.
Help on this plz?
To find the derivative of y with respect to x (dy/dx) and the second derivative (d2y/dx2), we can use the Fundamental Theorem of Calculus. Here's how you can proceed:
1. Start by noting that y is defined as a definite integral of a function with respect to t. Let's call this function f(t):
y = ∫[a,b] f(t) dt
2. In this specific problem, the function f(t) = 1/(t^2 + t + 1) and the limits of integration are a = 1 and b = 3x.
3. To find dy/dx, we'll differentiate both sides of the equation with respect to x:
d/dx(y) = d/dx(∫[a,b] f(t) dt)
4. According to the Fundamental Theorem of Calculus, when we differentiate an integral with respect to the variable in the limits of integration, we get the original function evaluated at the upper limit of integration multiplied by the derivative of the upper limit with respect to x, minus the original function evaluated at the lower limit of integration multiplied by the derivative of the lower limit with respect to x.
5. In this case, the limits of integration are a = 1 and b = 3x, so we have:
dy/dx = (f(b) * d/dx(b)) - (f(a) * d/dx(a))
6. Now we need to evaluate f(a), f(b), d/dx(a), and d/dx(b) to compute the derivative. Replace t with a and b in f(t) = 1/(t^2 + t + 1), and differentiate a and b with respect to x.
7. Once you have evaluated f(a), f(b), d/dx(a), and d/dx(b), substitute those values into the formula from step 5 to find dy/dx.
8. To find the second derivative, d2y/dx2, simply differentiate the result obtained in step 7 using standard rules of differentiation.
Following these steps will allow you to find the derivative (dy/dx) and the second derivative (d2y/dx2) of the given integral.