Fundamental Theorem of Calculus
The fundamental theorem of calculus is the statement that the two central operations of calculus, differentiation and integration, are inverse operations: if a continuous function is first integrated and then differentiated, the original function is retrieve..
First Fundamental Theorem of Integral Calculus
Let f(x) be a continuous function on the closed interval [a, b]. Let the area function A(x) be defined by th..
Let f(x) be a continuous function on the closed interval [a, b]. Let the area function A(x) be defined by th..Second Fundamental Theorem of Integral Calculus
Let f(x) be a continuous function defined on an interval [a,b]. between the limits a and b. This statement is also known as 'fundamental theorem of calculus'. We call b, the upper limit of x and a, the lower limit. If in place of F(x) we take F(x)+c as the value of the integral, we have =..
Let f(x) be a continuous function defined on an interval [a,b]. between the limits a and b. This statement is also known as 'fundamental theorem of calculus'. We call b, the upper limit of x and a, the lower limit. If in place of F(x) we take F(x)+c as the value of the integral, we have =..Suggested answer:
y 2 = 4ax (1) Differentiating with respect to x Substitute for 4a in (1), we get y - 2xy' = 0 Note that the given equation is differentiated only once to obtain the differential equation since it has only one consta..
y 2 = 4ax (1) Differentiating with respect to x Substitute for 4a in (1), we get y - 2xy' = 0 Note that the given equation is differentiated only once to obtain the differential equation since it has only one consta..Initial Value Problem
Because of these condition, the 2 n d order differential equation y''= 2 has particular solution x 2 + x + 2. The values f (0) = 2 and f '(0) = 1 are called initial values. The problem of finding the solution of a differential equation that satisfies these prescribed initial conditions is..
Because of these condition, the 2 n d order differential equation y''= 2 has particular solution x 2 + x + 2. The values f (0) = 2 and f '(0) = 1 are called initial values. The problem of finding the solution of a differential equation that satisfies these prescribed initial conditions is..Suggested answer:
Let y = vx v = log x +c Resubstituting for v, we get..
Let y = vx v = log x +c Resubstituting for v, we get..Suggested answer:
Let y = log u, u = tan t, t = x/2 From the above example, it is clear that chain rule can be extended. Let v = f(u), u = g(x), x = h(t) Then, we have..
Let y = log u, u = tan t, t = x/2 From the above example, it is clear that chain rule can be extended. Let v = f(u), u = g(x), x = h(t) Then, we have..Suggested answer:
The function where D x is a small change in x. = 2x - 9 f '(100) = 2 (100) - 9 = ..
The function where D x is a small change in x. = 2x - 9 f '(100) = 2 (100) - 9 = ..Suggested answer:
Let Differentiating with respect to x, we get ..
Let Differentiating with respect to x, we get ..Suggested answer:
Differentiating w.r.t t, we get Differentiating both sides with respect to x, we get ..
Differentiating w.r.t t, we get Differentiating both sides with respect to x, we get .. Result
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