4.9 FTC, part II

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In this section we learn the second part of the fundamental theorem and we use it to compute the derivative of an area function.

The Fundamental Theorem of Calculus, Part II

Area Functions

Area Function Let $f(t)$ be a continuous function on the interval $[a,b]$ and let $x$ be a number in the interval. Then, an area function is a function of the form $A(x) = \int _a^x f(t) \ dt.$

An area function gives us the net area between the curve $y = f(t)$ and the $t$ -axis over the interval $[a,x]$ . Net area means the area under the curve (where the function, $f(t)$ , is positive) minus the area above the curve (where the function, $f(t)$ , is negative).

Use the graph of $y = f(t)$ to find the values of the area function $A(x)= \int _0^x f(t) \ dt.$

$A(1) = \int _0^1 f(t) \ dt = \answer {2},$ $A(3) = \int _0^3 f(t) \ dt = \answer {4},$ $A(5) = \int _0^5 f(t) \ dt = \answer {2},$ $A(7) = \int _0^7 f(t) \ dt = \answer {0}.$

Use the graph of $y = f(t)$ to find the values of the area function $A(x)= \int _{-2}^x f(t) \ dt.$

From $-2$ to $2$ the curve is a semi-circle

$A(0) = \answer {\pi },$ $A(4) = \answer {2\pi - 2},$ $A(6) = \answer {2\pi -4},$ $A(7) = \answer {2\pi - 3.5}.$

FTC, Part II

The second part of the FTC tells us the derivative of an area function $A(x)$ .

Fundamental Theorem of Calculus, Part II If $f(x)$ is continuous on the closed interval $[a, b]$ then $A'(x) = \frac {d}{dx}\int _a^x f(t) \ dt = f(x),$ for any value of $x$ in the interval $[a, b]$ .

This conclusion establishes the theory of the existence of anti-derivatives, i.e., thanks to the FTC, part II, we know that every continuous function $f(x)$ has an anti-derivative.

Let $f(t) = \sin (t^2)$ . This function is continuous for all $x$ . Consider the function $A(x) = \int _a^x \sin (t^2) \ dt.$ By the FTC part II we can say that for any number $a$ and any value of $x$ , $\frac {d}{dx}\int _a^x \sin (t^2) \ dt = \sin (x^2).$ This example asserts that the continuous function $f(x) = \sin (x^2)$ has an anti-derivative. Finding another form- a familiar form- is an impossible task, despite the simple nature of the function itself. Thus the theory of anti-differentiation is much richer than the theory of differentiation, since such a simple function can be differentiated easily (using the chain rule).

Find the derivative of the function $A(x) = \int _0^x \cos (t) \ dt.$

By the FTC part II, $\frac {d}{dx}\int _0^x \cos (t) \ dt = \cos (x).$

Note that this equation can be verified easily by computing the integral and then taking the derivative of the result: $\int _0^x \cos (t) \ dt = \sin (t) \Big |_0^x = \sin (x) - \sin (0) = \sin (x)$ and $\frac {d}{dx} \sin (x) = \cos (x).$

$\frac {d}{dx} \int _0^x \sec ^3(t) \ dt = \answer {\sec ^3(x)}.$

$\frac {d}{dx} \int _1^x \ln ^2(t) \ dt = \answer {\ln ^2(x)}.$

Find the derivative of the function $A(x) = \int _1^x e^{2t}\sqrt {1+t^3} \ dt.$

By the FTC part II,

$A'(x) = \frac {d}{dx}\int _1^x e^{2t}\sqrt {1+t^3} \ dt = e^{2x}\sqrt {1+x^3}.$

Find the derivative of the function $A(x) = \int _x^0 f(t) \ dt.$ First, we rewrite the integral as $\int _x^0 f(t) \ dt =-\int _0^x f(t) \ dt.$ Then, by the FTC part II, $\frac {d}{dx}\int _x^0 f(t) \ dt = -\frac {d}{dx}\int _0^x f(t) \ dt =-f(x).$

$\frac {d}{dx} \int _x^0 e^t\cos (t) \ dt = \answer {-e^x\cos (x)}.$

Find the following derivative: $\frac {d}{dx} \int _0^{x^2} e^{2t}\sqrt {1+t^3} \ dt.$ To solve this problem, we will need to use the chain rule. Let $A(x) = \int _0^x e^{2t}\sqrt {1+t^3} \ dt.$ Then the function we wish to differentiate is $A(x^2)$ . By the chain rule, $\frac {d}{dx} A(x^2) = A'(x^2) \cdot 2x.$ By FTC, part II, $A'(x) = e^{2x}\sqrt {1+x^3}$ and substituting $x^2$ for $x$ , we have $A'(x^2) = e^{2x^2}\sqrt {1+(x^2)^3} = e^{2x^2}\sqrt {1+x^6}.$ Now multiplying by $2x$ gives the final answer: $\frac {d}{dx}\int _0^{x^2} e^{2t}\sqrt {1+t^3} \ dt = 2xe^{2x^2}\sqrt {1+x^6}.$

Find the following derivative: $\frac {d}{dx} \int _{2x}^0 \tan (3t) \ln (t^2) \ dt.$ To solve this problem, we will need to use the chain rule. Let $A(x) = \int _0^x \tan (3t) \ln (t^2) \ dt.$ Then the function we wish to differentiate is $-A(2x)$ . Note that the negative sign comes from switching the endpoints of integration. By the chain rule, $\frac {d}{dx} A(2x) = A'(2x) \cdot 2.$ By FTC, part II, $A'(x) = \tan (3x) \ln (x^2)$ and substituting $2x$ for $x$ , we have $A'(2x) = \tan (3(2x)) \ln ((2x)^2) = \tan (6x) \ln (4x^2).$ Now multiplying by $-2$ gives the final answer: $\frac {d}{dx} \int _{2x}^0 \tan (3t) \ln (t^2) \ dt = -2\tan (6x)\ln (4x^2).$

$\frac {d}{dx} \int _0^{x^3} t^4\cos (t) \ dt = \answer {3x^{14}\cos (x^3)}.$

$\frac {d}{dx} \int _{3x}^0 t^2\sin (t) \ dt = \answer {-27x^2\sin (3x)}.$

Here are some detailed, lecture style videos on area functions: