# integral addition with given area

• Nov 25th 2009, 03:02 PM
mybrohshi5
Let

$\displaystyle \int_{-4}^{3.5} f(x) dx =7, \ \int_{-4}^{-1.5} f(x) dx=10, \ \int_{1}^{3.5} f(x)dx =6.$

Find $\displaystyle \int_{-1.5}^{1} f(x)dx=$

and $\displaystyle \int_{1}^{-1.5} (7 f(x)- 10)dx=$

i found the first part to be -9 because 10+6+x=7 (solved for x)

i just cant get the second part. im just not exactly sure what it wants me to do.
thanks for any help =)
• Nov 25th 2009, 03:37 PM
Soroban
Hello, mybrohshi5!

Quote:

Let: . $\int_{\text{-}4}^{3.5}f(x)\,dx \:=\:7 \qquad \int_{\text{-}4}^{\text{-}1.5} f(x)\,dx\;=\;10 \qquad \int_{1}^{3.5} f(x)\,dx \:=\:6.$

Find $\int_{\text{-}1.5}^{1} f(x)\,dx$

and $\int_{1}^{\text{-}1.5} (7f(x) - 10)\,dx$

i found the first part to be -9 because 10+6+x=7 (solved for x)

i just cant get the second part. im just not exactly sure what it wants me to do.

The second one is confusing because of the switched limits.

I will switch them myself.

I have: . $\int^1_{\text{-}1.5}\bigg[7f(x) - 10\bigg]\,dx$

. . . . $=\; 7\underbrace{\int^1_{\text{-}1.5}f(x)\,dx}_{\text{This is -9}} \,-\,\int^1_{\text{-}1.5}10\,dx$

. . . . $= \;-63 - \bigg[10t\bigg]^1_{\text{-}1.5}$

. . . . $=\; -63 - \bigg[10 - (-15)\bigg]$

. . . . $=\; -63 - 25 \;=\;-88$

Therefore, their answer is: . $+88$

• Nov 25th 2009, 03:50 PM
mybrohshi5
oops i didnt even notice the a and b values switched. either way thank you for the great explanation