Area of the Region Between Two Curves

**MathGeek06** · August 10th 2008, 08:45 PM

Find the area of the region between y = x^2 and y = 4-3x from x = -1 to x = 4/3

**Chris L T521** · August 10th 2008, 08:52 PM

Originally Posted by MathGeek06

Find the area of the region between y = x^2 and y = 4-3x from x = -1 to x = 4/3

Did you graph the bounded region? This will help you set up the integral.

Recall that the area between two curves is $A=\int_a^b \left(f(x)-g(x)\right) \,dx$

I hope this helps!

--Chris

**MathGeek06** · August 10th 2008, 09:00 PM

Yes, I graphed the bounded region but when I set up my integrals, I got a negative answer, which clearly was incorrect.

I had an integral of ((4-3x) - x^3)) from x = -1 to x= 1 + the integral ((x^2) - (4-3x)) from x = 1 to x = 4/3.

**Chris L T521** · August 10th 2008, 09:24 PM

Originally Posted by MathGeek06

Yes, I graphed the bounded region but when I set up my integrals, I got a negative answer, which clearly was incorrect.

I had an integral of ((4-3x) - x^3)) from x = -1 to x= 1 + the integral ((x^2) - (4-3x)) from x = 1 to x = 4/3.

$\int_{-1}^1 (4-3x-x^2)\,dx+\int_1^{\frac{4}{3}}(x^2+3x-4)\,dx$

$\implies\left.\bigg[4x-\tfrac{3}{2}x^2-\tfrac{1}{3}x^3\bigg]\right|_{-1}^1+\left.\bigg[\tfrac{1}{3}x^3+\tfrac{3}{2}x^2-4x\bigg]\right|_1^{\frac{4}{3}}$

$\implies\bigg[\left(4-\tfrac{3}{2}-\tfrac{1}{3}\right)-\left(-4-\tfrac{3}{2}+\tfrac{1}{3}\right)\bigg]+\bigg[\left(\tfrac{64}{81}+\tfrac{8}{3}-\tfrac{16}{3}\right)-\left(\tfrac{1}{3}+\tfrac{3}{2}-4\right)\bigg]=\color{red}\boxed{\frac{1235}{162}}$

Does this make sense?

--Chris

Thread: Area of the Region Between Two Curves

LinkBack

Thread Tools

Display

Area of the Region Between Two Curves

Similar Math Help Forum Discussions

Finding the area of the region between the curves

Finding the area of a region bounded by THREE curves

One more area of region bounded by curves question...

Area of region bounded by these curves

area of the region that lies inside both curves

Search Tags