What is the area under the curve in the interval [0, 3] for the function #y=x^2+1#?

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Answer 1 · 2018-06-26T21:21:47

# =12 #

For this we must use integration

Here we are trying to find:

#int_0 ^3 (x^2 + 1) dx #

Use the reverse power rule:

#int x^n dx = x^(n+1)/(n+1) + "constant "#

# = [ x^(2+1) /(2+1) + x^1/(0+1) ]_0 ^3 #

# = [1/3x^3 +x ]_0 ^3 #

If we are trying to find #[F(x) ]_a ^b #

We just do #F(b) - F(a) #

#=>{ (1/3 * 3^3 ) +3 } - { (1/3*0^3) + 0 } #

# = 12#