Integrals - Area function

Find the derivative of the area function $A(x) = \int_{2}^{x} \sqrt{1 + t^2} dt$.

Given $A(x) = \int_{2}^{x} \sqrt{1 + t^2} dt$. Using the First Fundamental Theorem of Integral Calculus: $A'(x) = \frac{d}{dx} \int_{2}^{x} \sqrt{1 + t^2} dt$ $A'(x) = \sqrt{1 + x^2}$

According to the theorem, $\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)$. Here, $f(t) = \sqrt{1 + t^2}$, so we replace $t$ with $x$ to find the derivative.

If the area function is defined by $A(x) = \int_{1}^{x} (3t^2 + 2t) dt$, calculate the value of $A(2)$.

$A(2) = \int_{1}^{2} (3t^2 + 2t) dt$ Integrating the terms: $A(2) = \left[ \frac{3t^3}{3} + \frac{2t^2}{2} \right]_1^2$ $A(2) = [t^3 + t^2]_1^2$ Substituting the limits: $A(2) = (2^3 + 2^2) - (1^3 + 1^2)$ $A(2) = (8 + 4) - (1 + 1)$ $A(2) = 12 - 2 = 10$

To find $A(2)$, we substitute $x=2$ into the upper limit of the integral and evaluate the definite integral using the power rule of integration.

Verify the First Fundamental Theorem of Calculus for $f(x) = \cos x$ in the interval $[0, x]$.

First, find the area function $A(x)$: $A(x) = \int_{0}^{x} \cos t \, dt = [\sin t]_0^x = \sin x - \sin 0 = \sin x$ Now, find the derivative of $A(x)$: $A'(x) = \frac{d}{dx} (\sin x) = \cos x$ Since $A'(x) = f(x)$, the theorem is verified.

We explicitly calculated the area function by integrating $\cos t$ and then differentiated the result to show it returns the original function $f(x)$.

1. Set up the integral: $A(x) = \int_{0}^{x} (2t + 3) dt$.
2. Integrate with respect to $t$: $A(x) = [t^2 + 3t]_0^x$.
3. Substitute the limits: $A(x) = (x^2 + 3x) - (0^2 + 3(0)) = x^2 + 3x$.
4. Differentiate $A(x)$ with respect to $x$: $\frac{d}{dx}(x^2 + 3x) = 2x + 3$.
5. Compare with $f(x)$: Since $f(x) = 2x + 3$, we have verified $A'(x) = f(x)$.

The area under a linear function creates a trapezoidal region. The area function $x^2+3x$ calculates this area for any $x > 0$.

1. Evaluate the integral: $A(x) = [\ln |t|]_1^x = \ln x - \ln 1 = \ln x$.
2. Find the derivative using the First Fundamental Theorem: $A'(x) = f(x) = \frac{1}{x}$.
3. Calculate the value at $x = e$: $A'(e) = \frac{1}{e}$.

For the function $1/t$, the area starting from $t=1$ is the natural logarithm function. The slope of this area function at any point $x$ is $1/x$.