Question

In Exercises \(79-84,\) find \(F^{\prime}(x)\). $$ F(x)=\int_{x}^{x+2}(4 t+1) d t $$

Short answer

The derivative, \(F^{\prime}(x)\), of the given function is \(8\).

Step-by-step solution

Break down the given integral

\ Express the integral over the interval [x, x+2] as the difference of two integrals:\ \(F(x) = \int_{x}^{x+2} (4t + 1) dt = \int_{a}^{x+2} (4t + 1) dt - \int_{a}^{x} (4t + 1) dt\) \ Where a is a constant.

Apply the Fundamental Theorem of Calculus

We can make use of the Fundamental Theorem of Calculus which says that the derivative of an integral of a function is that original function. Therefore, we can directly find the derivative of the function with respect to \(x\).\ We differentiate each term individually:\ \((\int_{a}^{x+2} (4t+1) dt)^\prime = \frac{d}{dx} \int_{a}^{x+2} (4t+1) dt = (4(x+2) +1)\times \frac{d}{dx} (x+2) = (4x + 8 +1)\)\ \((\int_{a}^{x} (4t+1) dt)^\prime = \frac{d}{dx} \int_{a}^{x} (4t+1) dt = (4x +1)\)

Find the difference of the derivatives

The derivative \(F^{\prime}(x)\) is the difference of the derivatives found in step 2. Therefore,\ \(F^{\prime}(x) = (4x + 8 +1) - (4x +1) = 8\)