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

For this we must use integration

Here we are trying to find:

Use the reverse power rule:

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- What is #int (2x^3-3x^2-4x-3 ) / (-5x^2+ 2 x -4 )dx#?
- Let R be the region in the first quadrant bounded by the graphs of #(x^2/ 9) + (y^2 /81)=1# and #3x+y=9#, how do you find the area?
- What is the antiderivative of # (5/x)^2#?
- How do you find the definite integral for: #(x+1)^(2)dx# for the intervals #[-7, 0]#?
- How do you find #\int ( 4y + 3) ^ { - \frac { 1} { 2} } d y#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7