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.