Suppose F(t) is an antiderivative of t^{3}. Then F'(t)= t^3.
According to the Second Fundamental Theorem of Calculus,
x
∫ (t^3)dt = ???
1
Answer using the function F
F' = f(x)dx/dx - f(1)d1/dx
= t^3 - 0
= t^3
To evaluate the integral ∫ (t^3)dt, we can use the Second Fundamental Theorem of Calculus. According to the theorem, if F(t) is an antiderivative of f(t), then
∫[a to b] f(t) dt = F(b) - F(a).
In this case, F(t) is given as an antiderivative of t^3. So, we can directly integrate the function and use the theorem.
Since F(t) is an antiderivative of t^3, we have F'(t) = t^3.
Therefore, to find the value of the integral, we can use the Fundamental Theorem of Calculus as follows:
∫[a to b] t^3 dt = F(b) - F(a).
So, to find ∫ (t^3)dt, we need to substitute the limits of integration (a and b) into the function F(t) and then subtract the two values.
Please note that in your question, the limits of integration (a and b) are not provided. To evaluate the integral, you need to specify the limits of integration.