Question Statement

Evaluate the integral:

$\int (e^{4 x} - e^{- 1} + 1) d x$

Background and Explanation

This problem requires basic integration techniques, specifically the integration of exponential functions with linear arguments and the integration of constant terms. Recall that integration is linear, meaning we can integrate term by term, and constants can be factored out of the integral.

Solution

We evaluate the integral using the linearity property of integration, which allows us to integrate each term separately.

Let $I = \int (e^{4 x} - e^{- 1} + 1) d x$

Step 1: Apply the linearity of integration to split the integral into three separate terms. Note that $e^{- 1}$ is a constant (approximately $0.368$ ), so it can be factored out of the integral:

$I = \int e^{4 x} d x - e^{- 1} \int 1 d x + \int 1 d x$

Step 2: Evaluate each integral separately.

For the first term, using the rule $\int e^{a x} d x = \frac{e ^{a x}}{a}$ with $a = 4$ : $\int e^{4 x} d x = \frac{e ^{4 x}}{4}$

For the second term, since $e^{- 1}$ is a constant: $- e^{- 1} \int 1 d x = - e^{- 1} \cdot x = - e^{- 1} x$

For the third term: $\int 1 d x = x$

Step 3: Combine all terms and add the constant of integration $c$ :

$I = \frac{e ^{4 x}}{4} - e^{- 1} x + x + c$

This can also be written by factoring out $x$ from the last two terms as: $I = \frac{e ^{4 x}}{4} + x (1 - e^{- 1}) + c$

or equivalently: $I = \frac{e ^{4 x}}{4} + x (1 - \frac{1}{e}) + c$

Key Formulas or Methods Used

Linearity of Integration: $\int [f (x) + g (x)] d x = \int f (x) d x + \int g (x) d x$ and $\int k \cdot f (x) d x = k \int f (x) d x$ for constant $k$
Exponential Integration: $\int e^{a x} d x = \frac{e ^{a x}}{a} + c$ where $a \neq = 0$
Constant Integration: $\int k d x = k x + c$ where $k$ is any constant (including $e^{- 1}$ )

Summary of Steps

Split the integral using linearity: separate into $\int e^{4 x} d x$ , $- e^{- 1} \int 1 d x$ , and $\int 1 d x$
Integrate $e^{4 x}$ to get $\frac{e ^{4 x}}{4}$
Integrate the constant terms: $- e^{- 1}$ integrates to $- e^{- 1} x$ and $1$ integrates to $x$
Combine results and add the constant of integration $c$

Solution

We evaluate the integral using the linearity property of integration, which allows us to integrate each term separately.

Let

I = \int (e^{4 x} - e^{- 1} + 1) d x

Step 1: Apply the linearity of integration to split the integral into three separate terms. Note that

e^{- 1}

is a constant (approximately

0.368

), so it can be factored out of the integral:

I = \int e^{4 x} d x - e^{- 1} \int 1 d x + \int 1 d x

Step 2: Evaluate each integral separately.

For the first term, using the rule

\int e^{a x} d x = \frac{e ^{a x}}{a}

with

a = 4

\int e^{4 x} d x = \frac{e ^{4 x}}{4}

For the second term, since

e^{- 1}

is a constant:

- e^{- 1} \int 1 d x = - e^{- 1} \cdot x = - e^{- 1} x

For the third term:

\int 1 d x = x

Step 3: Combine all terms and add the constant of integration

c

I = \frac{e ^{4 x}}{4} - e^{- 1} x + x + c

This can also be written by factoring out

x

from the last two terms as:

I = \frac{e ^{4 x}}{4} + x (1 - e^{- 1}) + c

or equivalently:

I = \frac{e ^{4 x}}{4} + x (1 - \frac{1}{e}) + c

3.1 Q-7

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

3.1 Q-7

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps