Accumulation functions: the Fundamental Theorem, part 2

Three quick questions from earlier lessons.

$g(x) = \displaystyle\int_0^x 2t\,dt$ is the area under $y = 2t$ from 0 to $x$ , which is $x^2$ . What is $g'(x)$ ?

The area function grows at the rate of the height.

Fundamental Theorem, part 2: if

f

is continuous,

g(x) = \displaystyle\int_a^x f(t)\,dt

has

g'(x) = f(x)

. With a variable upper limit

u(x)

:

\dfrac{d}{dx}\displaystyle\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x)

.

$g(x) = \displaystyle\int_1^{x^3} \sqrt{t}\,dt$ . Find $g'(x)$ .

g(x) = F(x^3)

with

F' = \sqrt{t}

Upper limit u = x³.

g'(x) = \sqrt{x^3} \cdot 3x^2

Chain rule.

Your turn: $g(x) = \displaystyle\int_2^x (t^3 - t)\,dt$ .

g'(x) = x^3 - x

FTC part 2.

Here is the graph of $y = f(t)$ . Let $g(x) = \displaystyle\int_{-5}^{x} f(t)\,dt$ . Tap the point of this graph at the $x$ -value where $g$ has a local maximum.
Here is the graph of $y = f(t)$ . Let $g(x) = \displaystyle\int_{-5}^{x} f(t)\,dt$ . Tap the point of this graph at the $x$ -value where $g$ has a local maximum.

Three questions to check you have got it.

Accumulation functions

Watch out: the answer is the integrand, not its antiderivative.

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