#### To determine

**To evaluate:**

∫01x+1-x2dx

#### Answer

12+π4

#### Explanation

**1) Concept:**

Use the following formula to find the area of functions

Area of triangle=12·base·height

Area of circle=πr2

**2) Given:**

∫01x+1-x2dx

3) **Calculation:**

The given integral can be written as

∫01x+1-x2dx=∫01xdx+∫011-x2dx

Also we can write,

∫01x+1-x2dx=A1+A2

Where,

A1=∫01xdx and A2=∫011-x2dx

So we evaluate both integrals separately

First solve ∫01xdx

Draw the graph of y=x from 0, 1

A1=12·base·height

A1=12·1·1

A1=12

Now evaluate ∫011-x2dx

Draw the graph of y=1-x2 from 0, 1

From the graph,

y=1-x2 from 0, 1 is a quarter circle

So, A2 is one fourth area of an area of a circle and radius of the circle is 1

So the area is,

A2=14π12

A2=π4

So now integral becomes

∫01x+1-x2dx=12+π4

**Conclusion:**

Therefore,

∫01x+1-x2dx=12+π4