# Let f(x) = 1/x. 1. If x greater than 0 what values of a make \int _ { e } ^ { a } f ( t ) d t an...

## Question:

Let {eq}f(x) = 1/x {/eq} .

1. If {eq}x > 0 {/eq} what values of {eq}a {/eq} make {eq}\int _ { e } ^ { a } f ( t ) d t {/eq} an antiderivative of {eq}f {/eq} ?

2. What is the domain of {eq}F ( x ) = \int _ { 1 } ^ { x } 1 / t d t {/eq} ?

3.Find a formula for {eq}F(x) {/eq}.

4. Interpret {eq}F ( 5 ) = \int _ { 1 } ^ { 5 } 1 / t d t {/eq} geometrically.

## Antiderivatives

The antiderivative of {eq}\displaystyle f(x) {/eq} is a function {eq}\displaystyle F(x), \text{ such that }\frac{d}{dx}(F(x))=f(x). {/eq}

The general antiderivative function is {eq}\displaystyle F(x)+C \text{ because } \frac{d}{dx}(F(x)+C)=f(x), C -\text{ constant}. {/eq}

To solve a definite integral, we use the Fundamental Theorem of Calculus part II,

{eq}\displaystyle \int_a^b f(x)\ dx=F(x)\bigg\vert_a^b=F(b)-F(a),\\ {/eq} where {eq}\displaystyle F(x) \text{ is an antiderivative of } f(x): \frac{d}{dx}(F(x))=f(x). {/eq}

Part I of the Fundamental Theorem of Calculus states that

{eq}\displaystyle \frac{d}{dx}\left(\int_a^x f(t)\ dt \right)=f(x), \text{ where } a \text{ is any constant}. {/eq}

## Answer and Explanation:

1. If {eq}\displaystyle x>0 {/eq} then the function {eq}\displaystyle f(x)=\frac{1}{x} {/eq} is integrable.

The definite integral {eq}\displaystyle \int_e^a f(t)\ dt, \text{ where }a >e, \text{ needs to be determined } {/eq} is an antiderivative for {eq}\displaystyle f(x) {/eq} if

{eq}\displaystyle \frac{d}{dx}\left( \int_e^a f(t)\ dt\right)=f(x) \implies \boxed{a= x \text{ for the integral } \int_e^a f(t)\ dt \text{ to be an antiderivative of }f(x)}, {/eq}

by the Fundamental Theorem of Calculus, part I: {eq}\displaystyle \frac{d}{dx}\left( \int_b^x f(t)\ dt\right)=f(x), b\text{ being any constant }. {/eq}

2. The domain of the function {eq}\displaystyle F ( x ) = \int _ { 1 } ^ { x } \frac{1}{ t}\ d t {/eq}

is given by the values of {eq}\displaystyle x {/eq} that make the integral well defined.

Therefore, for all values {eq}\displaystyle x>1, \text{ the integral is well defined, } \boxed{\text{so the domain is }(1,\infty)}. {/eq}

3. To find a formula for {eq}\displaystyle F(x)=\int _ { 1 } ^ { x } \frac{1}{ t}\ d t, {/eq} we will integrate as below

{eq}\displaystyle F(x)=\int _ { 1 } ^ { x } \frac{1}{ t}\ d t=\ln|t|\bigg\vert_{t=1}^{t=x}=\ln|x|-\ln(1)=\boxed{\ln x}, \text{ because } x>0 \text{ and } ln(1)=0. {/eq}

4. The geometrical meaning of {eq}\displaystyle F ( 5 ) = \int _ { 1 } ^ { 5 } \frac{1}{ t} d t {/eq} is given by the geometrical meaning of a definite integral, which is

{eq}\displaystyle \boxed{\text{ the area under the positive function } f(t)=\frac{1}{t} \text{ between } t=1 \text{ and } t=5}. {/eq}

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