Fundamental Theorem of Calculus

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In section of the wiki, we will see how the two main branches of calculus( differentialand integral) calculus are related to each other. While the two might seem to beunrelated to each other, as one arose from the tangent problem, while the other arosefrom the area problem, we will see that the fundamental theorem of calculus does indeedcreate a link between the two.

ContentsThe first fundamental theorem of calculusSecond Fundamental theorem of calculus

The rst fundamental theorem of calculus

We have learned about indefinite integrals, which was the process of finding theantiderivative of a function. In contrast to the indefinite integral, the result of a definiteintegral will be a number, instead of a function. The definite integral of a function is thesigned area under the graph of the function, and is expressed in the form of:

\displaystyle{\int_a^b f(x)dx}.

Now suppose that we formed an area function S(x) in such a way that it is dependent onthe function f(x) as:

S(x)=\int _{ a }^{ x }{ f(t)dt }

Where f is continuous on the interval [a,b]. Know suppose we wanted to find the the rateof change of area with respect to x.

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We can see form the figure above that the area of the shaded region is equal to the areafrom a to x+\Delta x minus the area from a to x.

\Delta S=A(x+\Delta x)-A(x)

\frac{\Delta S}{\Delta x}=\frac{A(x+\Delta x)-A(x)}{\Delta x}

So the rate of change of area becomes:

S'(x)=\frac{dS}{dx}=\underset { x\rightarrow 0 }{ lim } \frac { s(x+\Delta x)-s(x) }{ \Deltax }

We know that there is an \overline{x} found between x and x+\delta x such that the areaof the shaded region is equal to f(\overline{x})\Delta x.

S'(x)=\underset { x\rightarrow 0 }{ lim } \frac { S(x+\Delta x)-S(x) }{ \Delta x }=\underset { x\rightarrow 0 }{ lim } \frac { f(\overline { x } )\Delta x }{ \Delta x } \\\quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad \quad =\underset { x\rightarrow 0 }{ lim } f(\overline {x } )\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad =f(x)\\ \\

The last step is true, as \Delta x\rightarrow 0, any thing found between x and x+\Deltax approaches x. So know we are ready to state the first fundamental theorem of calculus.

THEOREM

If f is continuous on [a,b] then the function defined by:

S(x)=\int _{ a }^{ x }{ f(t)dt }

is continuous on [a,b] and differentiable on (a,b) and S'(x)=f(x).

So basically integration is the opposite of differentiation. More clearly the firstfundamental theorem of calculus can written in Leibniz notation as:

\frac { d }{ dx } \int _{ a }^{ x }{ f(t)dt=f(x) } \\

EXAMPLE

Find the derivative of k(x)=\int _{ 2 }^{ x }{ ({ 4}^{ t }+t)dt } \\ .

The function f is continuous, so from the first fundamental theorem of calculus wehave:

k'(x)={ 4 }^{ x }+x

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EXAMPLE

What is the derivative of h(x)=\int _{ 2 }^{ { x }^{ 2 } }{ \frac { 1 }{ 1+{ t }^{ 2 } } dt }\\ .

We use The first fundamental theorem of calculus in accordance with the chain-rule to solve this.

Let u=x^{2}, then:

\frac { d }{ dx } \int _{ 2 }^{ { x }^{ 2 } }{ \frac { 1 }{ 1+{ t }^{ 2 } } dt= } \frac { d }{ du }\left[ \int _{ 1 }^{ u }{ \frac { 1 }{ 1+{ t }^{ 2 } } } dt \right] \cdot \frac { du }{ dx } \\\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad=\frac { 1 }{ 1+{ u }^{ 2 } } \cdot 4{ x }^{ 3 }\\ \\ \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad \quad \quad =\frac { 4{ x }^{ 3 } }{ 1+{ x }^{ 4 } } \\

EXAMPLE

Find the derivative of h(x)=\int _{ x }^{ 3x }{ sin\theta d\theta } .

We use the following property of integrals:

\int _{ b }^{ a }{ f(x)dx } =\int _{ b }^{ 0 }{ f(x)dx+\int _{ 0 }^{ a }{ f(x)dx } }

so

h'(x)=\frac { d }{ dx } \int _{ x }^{ 3x }{ sin\theta d\theta } \\ \\ \quad \quad \quad=\frac { d }{ dx } \int _{ x }^{ 0 }{ sin\theta d\theta +\frac { d }{ dx } \int _{ 0 }^{ 3x }{ sin\theta d\theta } } \\ \\ \quad \quad \quad =\frac { d }{ dx } -\int _{ 0 }^{ x }{sin\theta d\theta } +\frac { d }{ dx } \int _{ 0 }^{ 3x }{ sin\theta d\theta } \\ \\\quad \quad \quad \quad =-sinx+\frac { d }{ du } \int _{ 0 }^{ u }{ sin\thetad\theta } \cdot \frac { du }{ dx } \\ \\ \quad \quad \quad \quad \quad \quad\quad \quad =-sinx+3sin3x

Second Fundamental theorem of calculus

THEOREM

If f is a continuous function on [a,b] then

\int _{ a }^{ b }{ f(x)dx=F(b)-F(a) }

where F is the anti-derivative of f, that is F'=f.

PROOF

We know form the first fundamental theorem of calculus that if S(x)=\int _{ a }^{ x

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}{ f(t)dt } , then S is the anti-derivative of f or S'(x)=f(x). Let F be any otherantiderivative of f on the interval [a,b], then we know that F and S differ only by aconstant, that is:

F(x)=S(x)+c

for a

&=\frac{5}{4}-\left(-\frac{3}{4}\right)=2.\ _\square \end{align}

EXAMPLE

Evaluate \displaystyle{\int_{-3}^{2}(2x^2-3x+4)dx.}

Since \displaystyle{\int(2x^2-3x+4)dx=\frac{2}{3}x^3-\frac{3}{2}x^2+4x+C,} Wehave

\begin{align} \displaystyle{\int_{-3}^{2}(2x^2-3x+4)dx}&=\left[\frac{2}{3}x^3-\frac{3}{2}x^2+4x\right]_{-3}^{2}\\ &=\left(\frac{2}{3}\cdot2^3-\frac{3}{2}\cdot2^2+4\cdot2\right)-\left(\frac{2}{3}\cdot(-3)^3-\frac{3}{2}\cdot(-3)^2+4\cdot(-3)\right)\\ &=\frac{305}{6}.\ _\square \end{align}

EXAMPLE

Evaluate \displaystyle{\int_{4}^{9}\sqrt{x}dx.}

\begin{align} \int_{4}^{9}\sqrt{x}dx&=\int_4^9 x^{\frac{1}{2}}dx\\&=\left[\frac{2}{3}x^{\frac{3}{2}}\right]_4^9\\ &=\frac{2}{3}\cdot9^{\frac{3}{2}}-\frac{2}{3}\cdot4^{\frac{3}{2}}\\ &=\frac{38}{3}.\ _\square \end{align}

EXAMPLE

WHAT IS THE AREA OF THE SHADED FIGURE SHOWN ABOVE?We need to find the area under the curve y=\frac{1}{x} from x=\frac{1}{2} tox=\frac{5}{2}. Since \displaystyle{\int\frac{1}{x}dx=\ln x+C,} the area under thecurve is equal to:

\begin{align} \int_{\frac{1}{2}}^{\frac{5}{2}}\frac{1}{x}dx&=\left[\lnx\right]_{\frac{1}{2}}^{\frac{5}{2}}\\ &=\ln\frac{5}{2}-\ln\frac{1}{2}\\ &=\ln5.\_\square \end{align}

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Relevant For...

Calculus > Properties of IntegralsJEE Math (Beta) > Integration

EXAMPLE

Evaluate \displaystyle{\int_{-\frac{\pi}{2}}^{\frac{5}{6}\pi}\sin xdx.}

\begin{align} \int_{-\frac{\pi}{2}}^{\frac{5}{6}\pi}\sin xdx&=\left[-\cos x\right]_{-\frac{\pi}{2}}^{\frac{5}{6}\pi}\\ &=\frac{\sqrt{3}}{2}.\ _\square \end{align}

EXAMPLE

Find the area under the curve y=2x-x^2 from x=1 to x=2.

Since 2x-x^2\geq0 over the interval [1,2], the area under the curve y=2x-x^2 fromx=1 to x=2 is equal to \displaystyle{\int_1^2(2x-x^2)dx}. Since x^2-\frac{1}{3}x^3 isan antiderivative of 2x-x^2,

\begin{align} \int_1^2(2x-x^2)dx&=\left[x^2-\frac{1}{3}x^3\right]_1^2\\&=\left(2^2-\frac{1}{3}\cdot2^3\right)-\left(1^2-\frac{1}{3}\cdot1^3\right)\\&=\frac{2}{3}.\ _\square \end{align}

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