# The area of the region between the graph of y = 3x^{2} + 2x and the x-axis from x = 1 to x = 3 is

**Solution:**

Given, y = 3x^{2} + 2x

We have to find the area of the region between the graph y = 3x^{2} + 2x and the x-axis from x = 1 to x = 3.

Using integration,

Area = \(\int_{1}^{3}3x^{2}+2x\, dx \\\int x^{n}\, dx=\frac{x^{n+1}}{n+1}\)

We can use the second fundamental theorem of calculus.

\(\\Area \: =\: \left [ x^{3} +x^{2}\right ]from x=1\, to\, x=3 \\\int_{a}^{b}F(x)dx=f(b)-f(a)\)

Where, f’(x) = F(x)

Now, area = [(3)^{3 }+ (3)^{2}] - [(1)^{3 }+ (1)^{2}]

Area = (27 + 9) - (1 + 1)

Area = 36 - 2

Area = 34 square units

Therefore, the area of the region is 34 square units.

## The area of the region between the graph of y = 3x^{2} + 2x and the x-axis from x = 1 to x = 3 is

**Summary:**

The area of the region between the graph of y = 3x^{2} + 2x and the x-axis from x=1 to x = 3 is 34 square units.

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