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Fundamental Theorem of Calculus TAKE QUIZ

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

Contents 1 The first fundamental theorem of calculus 2 Second Fundamental theorem of calculus

The first fundamental theorem of calculus We have learned about indefinite integrals, which was the process of finding the antiderivative of a function. In contrast to the indefinite integral, the result of a definite integral will be a number, instead of a function. The definite integral of a function is the signed 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 on the 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 rate of 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 area from 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) }{ \Delta x} We know that there is an \overline{x} found between x and x+\delta x such that the area of 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+\Delta x 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 first fundamental 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 we have: 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 chainrule 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\theta d\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 other antiderivative of f on the interval [a,b], then we know that F and S differ only by a constant, that is: F(x)=S(x)+c for a<x
This theorem transforms the difficult problem of finding of evaluating definite integrals by calculation limits of sums, into a more easier problem of finding the antiderivative. So for example if asked to compute the integral \int _{ a }^{ b }{ f(x)dx } , we find the antiderivative of f(x) and compute their value at each end-point of the integral, and finally subtract them from each other. EXAMPLE

Evaluate \displaystyle{\int_0^1}x^2dx.

According to the fundamental theorem of calculus, we have \displaystyle{\int_0^1}x^2dx=F(1)-F(0), where F(x) is an antiderivative of x^2. Indefinite integration of x^2 gives \int x^2dx=\frac{1}{3}x^3+C, where C is the constant of integration. Hence we have F(1)-F(0)=\left(\frac{1}{3}\times1^3+C\right)-\left(\frac{1} {3}\times0+C\right)=\frac{1}{3}.\ _\square

Observe that the constant of integration C can be ignored, since it is to be eliminated regardless of its value. EXAMPLE

Find the area under the curve y=x^3+1 from x=-1 to x=1.

Since x^3+1\geq0 over the interval [-1,1], the area under the curve y=x^3+1 from x=-1 to x=1 is equal to \displaystyle{\int_{-1}^1(x^3+1)dx}. Since \frac{1}{4}x^4+x is an antiderivative of x^3+1, the area under the curve is: \begin{align} \int_{-1}^1(x^3+1)dx&=\left[\frac{1}{4}x^4+x\right]_{-1}^1\\ &=\left(\frac{1}{4}\cdot1^4+1\right)-\left(\frac{1}{4}\cdot(-1)^4+(-1)\right)\\ converted by Web2PDFConvert.com

&=\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,} We have \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} to x=\frac{5}{2}. Since \displaystyle{\int\frac{1}{x}dx=\ln x+C,} the area under the curve is equal to: \begin{align} \int_{\frac{1}{2}}^{\frac{5}{2}}\frac{1}{x}dx&=\left[\ln x\right]_{\frac{1}{2}}^{\frac{5}{2}}\\ &=\ln\frac{5}{2}-\ln\frac{1}{2}\\ &=\ln5.\ _\square \end{align}

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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 from x=1 to x=2 is equal to \displaystyle{\int_1^2(2x-x^2)dx}. Since x^2-\frac{1}{3}x^3 is an 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}

Relevant For... Calculus > Properties of Integrals JEE Math (Beta) > Integration

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