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Amy Ferguson Moncure

Calculus BC ยท Unit 6

The Fundamental Theorem and Accumulation Functions

Differentiate an integral with a moving limit, and evaluate an integral with an antiderivative.

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Key Ideas Easy Medium Hard

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

What to Remember

Evaluate with an antiderivative

If \(F\) is an antiderivative of \(f\), then

\(\int_a^b f(x)\,dx = F(b)-F(a)\)

An accumulation function differentiates back

If \(g(x)=\int_a^x f(t)\,dt\), then \(g'(x)=f(x)\).

\(g\) is increasing where \(f\) is positive, and \(g\) is concave up where \(f\) is increasing.

Variable limits need the chain rule

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

\(\dfrac{d}{dx}\int_{v(x)}^{u(x)} f(t)\,dt = f(u(x))\,u'(x)-f(v(x))\,v'(x)\)

A swapped lower limit brings a minus

If the upper limit is constant and the lower limit moves,

\(\dfrac{d}{dx}\int_{u(x)}^{a} f(t)\,dt = -f(u(x))\,u'(x)\)

Level 1

Easy

Either the upper limit is x, or the antiderivative is basic.

Example
Let \(g(x)=\int_1^x (t^2+1)\,dt\). Find \(g'(2)\).

\(g'(x)=x^2+1\)

\(g'(2)=2^2+1\)

\(5\)

Answer: \(5\)

1
Find \(\int_0^2 4x\,dx\).
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\(\int 4x\,dx = 2x^2\)

\(\left[2x^2\right]_0^2\)

\(2\cdot 4 - 0\)

\(8\)

Answer: \(8\)

2
Find \(\int_1^4 6\,dx\).
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\(\int 6\,dx = 6x\)

\(\left[6x\right]_1^4\)

\(24-6\)

\(18\)

Answer: \(18\)

3
Let \(g(x)=\int_0^x \sin t\,dt\). Find \(g'(\pi/2)\).
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\(g'(x)=\sin x\)

\(\sin(\pi/2)\)

\(1\)

Answer: \(1\)

4
\(f(3)=4\) and \(g(x)=\int_2^x f(t)\,dt\). Find \(g'(3)\).
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\(g'(x)=f(x)\)

\(g'(3)=f(3)\)

\(4\)

Answer: \(4\)

5
Find \(\int_0^1 e^x\,dx\).
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An antiderivative of \(e^x\) is \(e^x\).

\(\left[e^x\right]_0^1\)

\(e-1\)

Answer: \(e-1\)

Level 2

Medium

A limit is a function of x, or the accumulation function is read from a graph.

Example
Find \(\dfrac{d}{dx}\int_0^{x^2}\cos t\,dt\) at \(x=1\).

\(\dfrac{d}{dx}\int_0^{u(x)} f(t)\,dt = f(u(x))\,u'(x)\)

\(u(x)=x^2\) and \(u'(x)=2x\)

\(\cos(x^2)\cdot 2x\)

At \(x=1\): \(\cos 1 \cdot 2\).

\(2\cos 1\)

Answer: \(2\cos 1\)

1
The graph of \(f\) is the horizontal segment at height 2 from \(x=0\) to \(x=4\).

Graph of f. Horizontal segment at height 2.

Let \(g(x)=\int_0^x f(t)\,dt\). Find \(g(3)\) and \(g'(3)\).
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\(g(3)\) is the area of a 3 by 2 rectangle.

\(3\cdot 2\)

\(6\)

\(g'(3)=f(3)\)

\(2\)

Answer: \(g(3)=6\) and \(g'(3)=2\)

2
\(f(4)=-3\). Find \(\dfrac{d}{dx}\int_{2x}^{7} f(t)\,dt\) at \(x=2\).
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The upper limit is constant, so its derivative is 0.

\(\dfrac{d}{dx}\int_{2x}^{7} f(t)\,dt = -f(2x)\cdot 2\)

At \(x=2\), the lower limit is 4.

\(-f(4)\cdot 2\)

\(-(-3)\cdot 2\)

\(6\)

Answer: \(6\)

3
\(g(x)=\int_0^x f(t)\,dt\) and \(f\) changes from positive to negative at \(x=4\) only. Where does \(g\) have a local maximum?
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\(g'(x)=f(x)\)

\(g'\) changes from positive to negative at \(x=4\).

\(g\) changes from increasing to decreasing there.

The local maximum is at \(x=4\).

Answer: \(x=4\)

4
Find \(\dfrac{d}{dx}\int_1^{x^3} \sqrt{t+2}\,dt\).
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\(f(u(x))\,u'(x)\)

\(u(x)=x^3\) and \(u'(x)=3x^2\)

\(\sqrt{x^3+2}\cdot 3x^2\)

Answer: \(3x^2\sqrt{x^3+2}\)

5
\(g(x)=\int_0^x f(t)\,dt\). The values \(f(1)=2\), \(f(2)=0\), and \(f(3)=-4\) are given, and \(f\) is decreasing. On which interval is \(g\) concave down?
Show Answer

\(g''(x)=f'(x)\)

A decreasing \(f\) has \(f'(x)<0\).

So \(g''(x)<0\) on the whole interval where \(f\) is decreasing.

\(g\) is concave down on that entire interval.

Answer: on the interval where f is decreasing

Level 3

Hard

Both limits move, or a second derivative comes from the graph of f.

Example
Find the derivative of \(\int_x^{2x}(3t+1)\,dt\) at \(x=1\).

\(f(2x)\cdot 2 - f(x)\cdot 1\)

\((3\cdot 2x+1)\cdot 2 - (3x+1)\)

\((6x+1)\cdot 2 - 3x - 1\)

\(12x+2-3x-1\)

\(9x+1\)

At \(x=1\): \(10\).

Answer: \(10\)

1
Let \(g(x)=\int_2^{x^2} (t^2+t)\,dt\). Find \(g'(2)\).
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\(g'(x)=( (x^2)^2 + x^2 )\cdot 2x\)

\((x^4+x^2)\cdot 2x\)

At \(x=2\): inside is \(16+4=20\).

\(20\cdot 4\)

\(80\)

Answer: \(80\)

2
\(f\) is continuous, \(f(1)=5\), and \(f'(1)=-2\). Let \(g(x)=\int_0^x f(t)\,dt\). Find \(g''(1)\).
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\(g'(x)=f(x)\)

\(g''(x)=f'(x)\)

\(g''(1)=f'(1)\)

\(-2\)

Answer: \(-2\)

3
Find \(\dfrac{d}{dx}\int_{x}^{x^2} e^{t}\,dt\) at \(x=1\).
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\(e^{x^2}\cdot 2x - e^{x}\cdot 1\)

At \(x=1\):

\(e^{1}\cdot 2 - e\)

\(2e-e\)

\(e\)

Answer: \(e\)

4
\(g(x)=\int_1^x f(t)\,dt\), \(g(1)=0\), and \(f\) is positive on \((1, 5)\). Which statement is true?
A. \(g(4)>g(2)>0\)
B. \(g(4)
C. \(g(3)=0\)
D. \(g'(3)<0\)
Show Answer

\(g(1)=0\) and \(g'(x)=f(x)>0\) on \((1,5)\).

\(g\) is increasing on \([1,5]\), so the values stay positive and grow.

\(g(4)>g(2)>g(1)=0\)

Answer: A. \(g(4)>g(2)>0\)

5
Compute \(\int_0^3 |x-1|\,dx\) by splitting the integral where the inside of the absolute value changes sign.
Show Answer

\(|x-1|=1-x\) on \([0,1]\).

\(|x-1|=x-1\) on \([1,3]\).

\(\int_0^1 (1-x)\,dx = \left[x-\dfrac{x^2}{2}\right]_0^1\)

\(1-\dfrac{1}{2}\)

\(\dfrac{1}{2}\)

\(\int_1^3 (x-1)\,dx = \left[\dfrac{x^2}{2}-x\right]_1^3\)

\(\left(\dfrac{9}{2}-3\right) - \left(\dfrac{1}{2}-1\right)\)

\(\dfrac{3}{2} - \left(-\dfrac{1}{2}\right)\)

\(2\)

\(\dfrac{1}{2}+2\)

\(\dfrac{5}{2}\)

Answer: \(\dfrac{5}{2}\)

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