91Ó°ÊÓ

To find the area function A(x), we need to calculate the definite integral of the function f(t) from a to x. In this case, \(f(t)=t+5\) and \(a=-5\). Therefore, we will integrate the function with respect to t from -5 to x: $$A(x)=\int_{-5}^{x} (t+5) dt$$ Now let's find the integral of \((t+5)\) with respect to \(t\).

To integrate the function \((t+5)\) with respect to t, we need to find the antiderivative of the function: $$\int (t+5) dt = \frac{t^2}{2} + 5t + C$$ Now we will substitute the limits of integration to find the area function A(x).

Now we substitute the limits of integration (-5 and x) into the antiderivative: $$A(x) = \left(\frac{x^2}{2} + 5x + C\right) - \left(\frac{(-5)^2}{2} + 5(-5) + C\right)$$ Now simplify the expression and combine like terms to find A(x): $$A(x) = \frac{x^2}{2} + 5x -\left(\frac{25}{2} - 25\right) $$ $$A(x) = \frac{x^2}{2} + 5x + \frac{25}{2}$$ Now, we have found the area function A(x).

To draw the graph of the area function \(A(x)\), we must remember that the relationship between the function \(f(t)\) and its definite integral is related to the accumulation of the areas under the curve. The area function is a quadratic function given by \(A(x)=\frac{x^2}{2} + 5x + \frac{25}{2}\) and since this function has a positive leading coefficient it is an upward-facing parabola. The graph of the function \(f\) will be a straight line with a slope of \(1\) and a \(y\)-intercept at \(5\). So, for every point \((x, f(x))\) on the graph of \(f\), the area function \(A(x)\) represents the accumulated area underneath the curve up to that point. Now, let's verify that the derivative of the area function equals f(x).

To verify that \(A'(x) = f(x)\), we need to find the derivative of the area function \(\frac{x^2}{2} + 5x + \frac{25}{2}\) with respect to x: $$A'(x) = \frac{d}{dx} \left(\frac{x^2}{2} + 5x + \frac{25}{2}\right)$$ Now, using the power rule and constant rule, we get: $$A'(x) = x + 5$$

From Step 5 we found out that \(A'(x) = x + 5\). The original function, \(f(t)\), is given by \(f(t) = t + 5\). Since we are using 'x' as our variable instead of 't' in \(A'(x)\) notation, it is equivalent to write \(f(x) = x + 5\). Therefore, we have verified that: $$A'(x)=f(x)$$