Question

Find the integration of the given function within the given limits: $\int_0^4 {(x + {e^{2x}})dx}$

Answer 1

Hint : Here, we calculate the integral of ${e^{2x}}$ separately by substituting $2x = u$ and using the u-substitution method. Then we add the integrals of $x$ and ${e^{2x}}$ and apply the limits 0 and 4 to get the answer.

Complete step-by-step answer :
We are given a function $x + {e^{2x}}$ and we are asked to find its definite integral with limit ranging from 0 to 4.
We can see that the given function is of the form $f + g$ where $f$ and $g$ are two different functions.
Therefore, we can apply the sum rule of integration which is given as follows:
$\int {f(x) \pm g(x)} dx = \int {f(x)dx} \pm \int {g(x)dx}$
Comparing the given function $x + {e^{2x}}$ with the sum rule, we can take $f(x) = x$ and $g(x) = {e^{2x}}$ .
Thus, on substitution, we get
$\int_0^4 {(x + {e^{2x}})dx} = \int_0^4 {xdx + \int_0^4 {{e^{2x}}dx} }$
Let’s number this equation as follows: $\int_0^4 {(x + {e^{2x}})dx} = \int_0^4 {xdx + \int_0^4 {{e^{2x}}dx} } .....(1)$
We will compute the integral $\int {{e^{2x}}dx}$ separately by using the u-substitution method.
Let $2x = u$ . Then, we get $\dfrac{1}{2}u = x$ .
Differentiating both sides of the equation $\dfrac{1}{2}u = x$ with respect to $x$ , we get
$\dfrac{1}{2}du = dx$ .
Therefore, on substitution, we have
$\int {{e^{2x}}dx} = \int {\dfrac{1}{2}{e^u}du}$
We take the constant out in this step.
$\int {{e^{2x}}dx} = \dfrac{1}{2}\int {{e^u}du}$
We know that $\int {{e^u}du} = {e^u} + C$ where C is a constant.
Using this, we get$\int {{e^{2x}}dx} = \dfrac{1}{2}{e^u}$
Now, let’s substitute $u = 2x$ . Then we get$\int {{e^{2x}}dx} = \dfrac{1}{2}{e^{2x}}.....(2)$
On substituting the integral obtained (2) in equation (1), we get
$\int_0^4 {(x + {e^{2x}})dx} = \int_0^4 {xdx + \left[ {\dfrac{1}{2}{e^{2x}}} \right]} _0^4$
We know that, $\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$.
Using this in the above equation we get

$$\int_0^4 {(x + {e^{2x}})dx} \\\ = \left[ {\dfrac{{{x^2}}}{2}} \right]_0^4 + \left[ {\dfrac{1}{2}{e^{2x}}} \right]_0^4 + C \\\ = \dfrac{{{4^2}}}{2} - \dfrac{{{0^2}}}{2} + \dfrac{1}{2}{e^{2 \times 4}} - \dfrac{1}{2}{e^{2 \times 0}} + C \\\ = 8 - 0 + \dfrac{1}{2}{e^8} - \dfrac{1}{2}{e^0} \\\$$

We know that ${e^0} = 1$ .
Therefore, we have

$$\int_0^4 {(x + {e^{2x}})dx} \\\ = 8 + \dfrac{1}{2}{e^8} - \dfrac{1}{2} \\\ = \dfrac{1}{2}(16 - 1 + {e^8}) \\\ = \dfrac{1}{2}(15 + {e^8}) \\\$$

Hence $\dfrac{1}{2}(15 + {e^8})$ is the required answer.

Note : It is a common tendency among students to substitute $du$ for $dx$ while using the u-substitution method. However, this will lead you to a wrong answer. The operation of integration, up to an additive constant, is the inverse of the operation of differentiation.