How to solve indefinite integral of e rise to four x minus 5 by dx ?
Short answer: indefinite integral of e rise to four x minus 5 by dx is (1/4)e^(4x-5)+C. ∫ (e^(4x-5))dx is not particulary hard integral. You can solve it in 7 easy steps. We will walk you through and explain everything. Let's start.
Table of contents
- Required assumtions
- Step by step solution of ∫ (e^(4x-5))dx
- What is indefinite integral of e rise to four x minus 5 by dx?
- Full video how to solve ∫ (e^(4x-5))dx
- Dictionary
- Want to solve next one?
Required assumtions
Usually, we have some additional info about function f of (x). In our case:
- f(x) belongs to real numbers
- f(x) is integrable in that domain
Step by step solution of ∫ (e^(4x-5))dx
We will solve ∫ (e^(4x-5))dx in 7 easy steps. Let's get started
Step 1
From symbol dx we know that differential of variable x indicates that the variable of integration is x. In that case, to solve this integral we may apply so called u−substitution method. We do it in order to modify the integral in the way that enable using some of formulas for finding integrals. In our case we substitute expression in power.
Step 2
We say u is equal to 4x minus 5. We also have to know what is the derivative of u. So we write derivative of u equals to d 4x minus 5. From rules that we can apply to derivatives, we know that derivative of 4x minus 5 equals derivative of 4x minus derivative of 5. Thus, we have du equals d 4x minus d5. We also know that derivative of constant is 0. Consequently, our derivative of 5 is also 0. And we’ve got du is equal to 4 dx. But what we want to have is expression that we insert to our main solution instead of dx. So, after division of both sides by 4 we’ve got du divided by 4 is equal to dx that we were searching for now.
Step 3
Let’s make use of our u. We insert u instead of 4x minus 5 in power of our main equation and instead of dx we now use expression with our substituted variable u which is du divided by 4. We have now integral of e rise to u by du divided by 4.
Step 4
Next, we try to modify our main equation to the form which enables using some known formula for solving integral. First, knowing that integral of a multiplied by s by ds, where a is some constant equals a multiplied by integral of s by ds, we may apply it to our equation. In our case s is e rise to u and a is one-quarter.
Step 5
Perfect, now we can apply formula which says integral of e rise to s by ds equals e rise to s. We also must add constant C to make sure that we have all our solution taken into account as our solution make whole class of function not a single one. In our case s is u
Step 6
and we’ve got in our main equation one-quarter multiplied by e rise to u plus constant C.
Step 7
But it is not the end of our solution. We have to come back to our problem, and it was formulated using variable x. We use u-substitution and now we have to come back to x. So instead of u we now insert 4x minus 5. We’ve got e rise to 4x minus 5 plus constant C as a result of our problem.
What is indefinite integral of e rise to four x minus 5 by dx?
We finally did it: ∫ (e^(4x-5))dx=(1/4)e^(4x-5)+C
Dictionary
Integration (antidifferentiation)
Computation (process of finding) of an integral, opposite process to differentiation.Integrand
Function placed between sign of integral and differential of variable of integration e.g. $${{ \int f(x)dx}}$$,where:
$${{ \int }}$$- integration operator,
f(x) – integrand,
dx- differential of variable of integration x
Integrable function
Function that integral over its domain is finite.Indefinite integral
Represents a class of primitive functions whose derivative is the integrand e.g. $${{\int f(x)dx=F(x)+C \Leftrightarrow F’(x)=f(x)}}$$$${{C=const. }}$$,
$${{f, F,C \in R }}$$
R-real numbers
$${{\int f(x)dx}}$$ - indefinite integral of function f(x) by dx,
$${{F(x)+C}}$$ – a class of primitive functions that $${{F’(x)=f(x) }}$$,
F(x) - primitive function, usually written in capital letters,
R-real numbers.
Integration by substitution, substitution method, u-substitution
Method used to solve integral when an integral contains some function and its derivative. The function we sat as equal to u and we rewrite the integral using a new variable u.$${{\varphi: x \rightarrow \varphi (x)=u, x \in X, u \in U}}$$
and
$${{f: u \rightarrow f(u), u \in U, f(u) \in \varphi }}$$
$${{\varphi}}$$ is injective and differentiable function in X
f is continuous in U
then, $${{ \int f(\varphi (x)) \varphi’(x)dx=f(u) du}}$$ $${{\int f(\varphi (x))\varphi’(x)dx=\Bigg|\matrix{u= \varphi (x) \cr d(u)= d(\varphi (x))\cr du= \varphi ’(x) dx}\Bigg| =\int f(u) du}}$$
Function
Function specified on a set X and having values in set Y is an assignment each element of set X specifically one element in set Y.$${{f: X \rightarrow Y}}$$
f-function name,
X-set of elements of function f, domain of a function f
Y-set of function values of function f, codomain of a function f
$${{x\in X, y \in Y}}$$
$${{y=f(x) f: x\rightarrow y}}$$
$${{y=y(x), }}$$
y(x)-vales of the function named y,
x-independent variable,
y-dependent variable.
Derivative, derivative of a function
If $${{lim {\Delta f \over \Delta x }}}$$exists (and is finite), then this limit is named$${{ \Delta x\rightarrow 0}}$$
derivative of a function f(x) in a point x0 and use symbol f’(x).
$${{ f’(x_0) {\buildrel\rm def\over=} \lim_{\Delta x\to0} {\Delta f \over \Delta x}}}$$
Derivative function f’(x)
Derivative function of function f(x) is a function that values in a point x_0 are equal to derivative of this function in the point x_0.f’(x_0) - derivative of a function f(x) in a point x_0 (a number, because it is a specifically determined limit of a function),
f’(x)-derivative function.
Differentiable function
f(x) is differentiable in a point x_0 means that there exists f’(x_0 )-f(x) is differentiable in (a,b) means there exists f’(x), x $${{\in(a,b) }}$$
-f(x) is differentiable in
