Can you evaluate the integral by interpreting it in terms of areas?

Question

Can you evaluate the integral by interpreting it in terms of areas?

1 Answer

Impact of this question

11919 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-28T16:45:14

$\int_{- 4}^{3} 1 - x d x = \frac{21}{2}$ $int_(-4)^3 1 -xdx = 21/2$

Explanation:

Let's sketch the graph of $f (x) = 1 - x$ $f(x) = 1 - x$ . Add the vertical lines $x = - 4$ $x =-4$ and $x = 3$ $x = 3$ .

enter image source here

Integrating is just finding the area under a curve, so since we have two triangle, we just find their areas, add them up and we're all done.

The first triangle has base length of $5$ $5$ units and height $5$ $5$ units. The area is therefore $\frac{5 \cdot 5}{2} = \frac{25}{2}$ $(5 * 5)/2 = 25/2$ .

The second triangle has base length $2$ $2$ and height $- 2$ $-2$ . Therefore the area will be $\frac{2 \cdot - 2}{2} = - 2$ $(2 * -2)/2 = -2$

Therefore $\int_{- 4}^{3} 1 - x d x = \frac{25}{2} + (- 2) = \frac{21}{2}$ $int_(-4)^3 1 - x dx = 25/2 + (-2) = 21/2$

Hopefully this helps!

Science

Math

Humanities

... and beyond

Can you evaluate the integral by interpreting it in terms of areas?

1 Answer

Explanation:

Related questions

Impact of this question