evaluate the following expression at z=5
d/dz ¡Ò[z to 2] (3+x^2)dx
To evaluate the given expression, we need to find the derivative of the integral with respect to z, and then evaluate it at z = 5.
Let's break down the steps to solve this problem:
Step 1: Evaluate the integral.
The integral ¡Ò[z to 2] (3+x^2)dx means that we are integrating the function (3+x^2) with respect to x from the lower limit z to the upper limit of 2. To evaluate this integral, we need to find the antiderivative of the function (3+x^2) with respect to x.
The antiderivative of (3+x^2) with respect to x is (3x + (x^3)/3).
Now, we evaluate the antiderivative at the upper limit (2) and subtract the value of the antiderivative at the lower limit (z):
F(2) - F(z) = [(3(2) + (2^3)/3)] - [(3z + (z^3)/3)] = (6 + 8/3) - (3z + z^3/3)
The result is (14/3) - (3z + z^3/3).
Step 2: Take the derivative with respect to z.
To find d/dz of the result obtained in Step 1, we differentiate with respect to z:
d/dz[(14/3) - (3z + z^3/3)] = -3 - (z^2/3).
Step 3: Substitute z = 5.
Finally, we substitute z = 5 into the derivative we obtained in Step 2:
d/dz[(14/3) - (3z + z^3/3)]|z=5 = -3 - (5^2/3)
= -3 - 25/3
= -9 - 25/3
= -9 - 8 1/3
= -17 1/3
Therefore, the value of the expression at z = 5 is -17 1/3.