To find the value of f'(π/2), we need to differentiate the function f(x) = ∫(0 to x) e^(t^2) sin(t^3) dt with respect to x and then evaluate it at x = π/2.
Let's denote the integral as F(x):
F(x) = ∫(0 to x) e^(t^2) sin(t^3) dt
To differentiate F(x) with respect to x, we can use the Fundamental Theorem of Calculus. According to this theorem, if we have a function F(x) defined as an integral from a constant lower limit (in this case, 0) to a variable upper limit (in this case, x), then its derivative is given by the integrand evaluated at the upper limit (x):
F'(x) = d/dx [∫(0 to x) e^(t^2) sin(t^3) dt]
= e^(x^2) sin(x^3)
Now, we can evaluate F'(π/2) by substituting x = π/2 into the expression for F'(x):
F'(π/2) = e^((π/2)^2) sin((π/2)^3)
= e^(π^2/4) sin(π^3/8)
Therefore, the value of f'(π/2) is e^(π^2/4) sin(π^3/8).
The short answer is that without performing the calculations or approximations, we cannot determine the value of f'(π/2) directly. It requires evaluating the integral and performing numerical computations to approximate the derivative at that specific point.
Step-by-step explanation:
To find the value of f'(π/2), we need to take the derivative of the function f(x) and then evaluate it at x = π/2.
Let's first find the derivative of f(x) using the Fundamental Theorem of Calculus. According to the theorem, if F(x) is the antiderivative of f(x), then the derivative of the integral from a constant lower limit (in this case, 0) to a variable upper limit (in this case, x) is given by f(x).
In our case, f(x) = ∫(0 to x) e^(t^2) sin(t^3) dt. We can rewrite it as follows:
f(x) = F(x) - F(0),
where F(x) is the antiderivative of e^(t^2) sin(t^3).
Now, let's find the antiderivative F(x):
F(x) = ∫(0 to x) e^(t^2) sin(t^3) dt.
To evaluate this integral, we need to use techniques such as integration by parts and substitution. However, the resulting expression for F(x) would be quite complex and difficult to work with.
Instead, we can use a numerical approach to find the value of f'(π/2). We'll approximate the derivative using a small change in x and calculate the corresponding change in f(x).
Let's choose a small value for Δx, say Δx = 0.001. Then, we can calculate f(π/2 + Δx) and f(π/2) using the given integral:
Finally, we can approximate the derivative f'(π/2) as:
f'(π/2) ≈ (f(π/2 + Δx) - f(π/2)) / Δx.
By substituting the values of f(π/2 + Δx) and f(π/2) obtained from the numerical integration, and Δx = 0.001, we can calculate the approximate value of f'(π/2).
Please note that this numerical approach provides an approximation, and the result may not be exact.
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Answers & Comments
Answer:
To find the value of f'(π/2), we need to differentiate the function f(x) = ∫(0 to x) e^(t^2) sin(t^3) dt with respect to x and then evaluate it at x = π/2.
Let's denote the integral as F(x):
F(x) = ∫(0 to x) e^(t^2) sin(t^3) dt
To differentiate F(x) with respect to x, we can use the Fundamental Theorem of Calculus. According to this theorem, if we have a function F(x) defined as an integral from a constant lower limit (in this case, 0) to a variable upper limit (in this case, x), then its derivative is given by the integrand evaluated at the upper limit (x):
F'(x) = d/dx [∫(0 to x) e^(t^2) sin(t^3) dt]
= e^(x^2) sin(x^3)
Now, we can evaluate F'(π/2) by substituting x = π/2 into the expression for F'(x):
F'(π/2) = e^((π/2)^2) sin((π/2)^3)
= e^(π^2/4) sin(π^3/8)
Therefore, the value of f'(π/2) is e^(π^2/4) sin(π^3/8).
Answer:
The short answer is that without performing the calculations or approximations, we cannot determine the value of f'(π/2) directly. It requires evaluating the integral and performing numerical computations to approximate the derivative at that specific point.
Step-by-step explanation:
To find the value of f'(π/2), we need to take the derivative of the function f(x) and then evaluate it at x = π/2.
Let's first find the derivative of f(x) using the Fundamental Theorem of Calculus. According to the theorem, if F(x) is the antiderivative of f(x), then the derivative of the integral from a constant lower limit (in this case, 0) to a variable upper limit (in this case, x) is given by f(x).
In our case, f(x) = ∫(0 to x) e^(t^2) sin(t^3) dt. We can rewrite it as follows:
f(x) = F(x) - F(0),
where F(x) is the antiderivative of e^(t^2) sin(t^3).
Now, let's find the antiderivative F(x):
F(x) = ∫(0 to x) e^(t^2) sin(t^3) dt.
To evaluate this integral, we need to use techniques such as integration by parts and substitution. However, the resulting expression for F(x) would be quite complex and difficult to work with.
Instead, we can use a numerical approach to find the value of f'(π/2). We'll approximate the derivative using a small change in x and calculate the corresponding change in f(x).
Let's choose a small value for Δx, say Δx = 0.001. Then, we can calculate f(π/2 + Δx) and f(π/2) using the given integral:
f(π/2 + Δx) = ∫(0 to π/2 + Δx) e^(t^2) sin(t^3) dt,
f(π/2) = ∫(0 to π/2) e^(t^2) sin(t^3) dt.
Finally, we can approximate the derivative f'(π/2) as:
f'(π/2) ≈ (f(π/2 + Δx) - f(π/2)) / Δx.
By substituting the values of f(π/2 + Δx) and f(π/2) obtained from the numerical integration, and Δx = 0.001, we can calculate the approximate value of f'(π/2).
Please note that this numerical approach provides an approximation, and the result may not be exact.
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