Antiderivatives and Integrals

Area Under a Curve

Find the area below the curve $\displaystyle f(x) = \sin(x)$ and above the x-axis between $\displaystyle x = {\pi \over 2}$ and $\displaystyle x = \pi$ .

\displaystyle f(x)=\sin x

What if we have a more interesting situation where many curves are involved? For instance, how do we find the area between two curves $\displaystyle f$ and $\displaystyle g$ ?

\displaystyle f(x)=\sqrt{1-x}

\displaystyle g(x)=2

\displaystyle \text{Area under Upper} - \text{Area under Lower} = \text{Area expected}

Calculate the area bounded by $\displaystyle y = 2x + 1$ and $\displaystyle y = x^2 - 2x - 3$ .

\displaystyle y=2x+1

\displaystyle y=x^2-2x-3

Calculate the area bounded by $\displaystyle y = -x$ , $\displaystyle y = \sqrt{x}$ , and $\displaystyle y = -x^2 + 2$ between $\displaystyle x = 0$ and $\displaystyle x = 2$ .

\displaystyle y=-x

\displaystyle y=\sqrt{x}

\displaystyle y=-x^2+2

Find the area bounded by $\displaystyle y^2 = x + 4$ and $\displaystyle y = {1 \over 2}x + {1 \over 2}$ .

\displaystyle x=y^2-4

\displaystyle x=2y-1

Choose a pair of curves from the source and fill the region with midpoint rectangles. Vertical slices use upper minus lower; horizontal slices use right minus left. Increase the number of slices and compare the approximation with the exact area.

Exercise and slice directionNumber of slices n8

\displaystyle y=2x+1

\displaystyle y=x^2-2x-3

\displaystyle A=\int_{2-2\sqrt2}^{2+2\sqrt2}((2x+1)-(x^2-2x-3))\,dx=\frac{64\sqrt2}{3}

Slice area sum: 30.405592 · Exact area: 30.169889 · Absolute error: 0.235702

QUESTION

For horizontal slices between

\displaystyle x=y^2-4

and

\displaystyle x=2y-1

, which integral gives the area?

◀ Previous SectionDefinite Integral Course CompletedBack to Syllabus

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Antiderivatives and Integrals

Area Under a Curve

Find the area below the curve $\displaystyle f(x) = \sin(x)$ and above the x-axis between $\displaystyle x = {\pi \over 2}$ and $\displaystyle x = \pi$ .

\displaystyle f(x)=\sin x

What if we have a more interesting situation where many curves are involved? For instance, how do we find the area between two curves $\displaystyle f$ and $\displaystyle g$ ?

\displaystyle f(x)=\sqrt{1-x}

\displaystyle g(x)=2

\displaystyle \text{Area under Upper} - \text{Area under Lower} = \text{Area expected}

Calculate the area bounded by $\displaystyle y = 2x + 1$ and $\displaystyle y = x^2 - 2x - 3$ .

\displaystyle y=2x+1

\displaystyle y=x^2-2x-3

Calculate the area bounded by $\displaystyle y = -x$ , $\displaystyle y = \sqrt{x}$ , and $\displaystyle y = -x^2 + 2$ between $\displaystyle x = 0$ and $\displaystyle x = 2$ .

\displaystyle y=-x

\displaystyle y=\sqrt{x}

\displaystyle y=-x^2+2

Find the area bounded by $\displaystyle y^2 = x + 4$ and $\displaystyle y = {1 \over 2}x + {1 \over 2}$ .

\displaystyle x=y^2-4

\displaystyle x=2y-1

Exercise and slice directionNumber of slices n8

\displaystyle y=2x+1

\displaystyle y=x^2-2x-3

\displaystyle A=\int_{2-2\sqrt2}^{2+2\sqrt2}((2x+1)-(x^2-2x-3))\,dx=\frac{64\sqrt2}{3}

Slice area sum: 30.405592 · Exact area: 30.169889 · Absolute error: 0.235702

QUESTION

For horizontal slices between

\displaystyle x=y^2-4

and

\displaystyle x=2y-1

, which integral gives the area?

◀ Previous SectionDefinite Integral Course CompletedBack to Syllabus

Area Under a Curve