# Evaluate the indefinite integral \int_0^4 (2e^x+3 \sin x) dx

## Question:

Evaluate the indefinite integral {eq}\int_0^4 (2e^x+3 \sin x) dx {/eq}

## Indefinite Integral:

The indefinite integral of a function is the expression which when differentiated, gives the function itself. There are various formulas given which can be directly applied to find the indefinite integral of a function. Some of the formulas are:

{eq}\displaystyle { \int e^x \ dx = e^x \\ \int \sin x = - \cos x } {/eq}

When the indefinite integration is done over an interval then it is called Definite integration. The definite integration gives us the area of the function curve under the function curve, x-axis and upper and lower limits.

Given:

{eq}\displaystyle { \int_0^4 (2e^x+3 \sin x) dx } {/eq}

On integrating it we get:

{eq}\displaystyle { = \left [ (2e^x + 3 (-\cos x) \right ]_0^4 \\ = \left [ 2e^x - 3 \cos x \right ]_0^4 } {/eq}

which is the indefinite integral of the given expression. Now if we apply the limits, we get the definite integral of the given function over given interval:

{eq}\displaystyle { = \left [ 2e^4 - 3 \cos 4 \right ] - \left [ 2e^0 - 3 \cos 0 \right ] \\ = 2e^4 - 3 \cos 4 - 2 + 3 \\ = 1 + 2e^4 - 3 \cos 4 \\ \approx 112.16 \ \text{Units} } {/eq} 