1.2 The definite integral; Questions 88-93

[Eigenvalue]

Suppose that [tex]\int\limits_{0}^{4}[/tex]f(x)dx=5, [tex]\int\limits_{0}^{2}[/tex]f(x)dx=-3, [tex]\int\limits_{0}^{4}[/tex]g(x)dx=-1, and [tex]\int\limits_{0}^{2}[/tex]g(x)=2. In the following exercises, compute the integrals

88. [tex]\int\limits_{0}^{4}[/tex](f(x)+g(x))dx
=[tex]\int\limits_{0}^{4}[/tex]f(x)dx+[tex]\int\limits_{0}^{4}[/tex]g(x)
=5+(-1)
=4

88[tex]\frac{1}{2}[/tex]. [tex]\int\limits_{0}^{2}[/tex](f(x)+g(x))dx
=[tex]\int\limits_{0}^{2}[/tex]f(x)dx+[tex]\int\limits_{0}^{2}[/tex]g(x)dx
=-3+2
=-1

89.[tex]\int\limits_{2}^{4}[/tex](f(x)+g(x))dx
=[tex]\int\limits_{2}^{0}[/tex](f(x)+g(x))+[tex]\int\limits_{0}^{4}[/tex](f(x)+g(x)
=-[tex]\int\limits_{0}^{2}[/tex](f(x)+g(x))dx+[tex]\int\limits_{0}^{4}[/tex](f(x)+g(x))dx
=-(-3+2)+(5-1)
=-1+4
=3

90. [tex]\int\limits_{0}^{2}[/tex](f(x)-g(x))dx
=[tex]\int\limits_{0}^{2}[/tex]f(x)dx-[tex]\int\limits_{0}^{2}[/tex]g(x)dx
=-3-2
=-5

91. [tex]\int\limits_{2}^{4}[/tex](f(x)-g(x))dx
=[tex]\int\limits_{2}^{0}[/tex](f(x)-g(x))dx+[tex]\int\limits_{0}^{4}[/tex](f(x)-g(x))dx
=-[tex]\int\limits_{0}^{2}[/tex](f(x)-g(x))dx+[tex]\int\limits_{0}^{4}[/tex](f(x)-g(x))dx
=-(-3-2)+(5+1)
=5+6
=11

92. [tex]\int\limits_{0}^{2}[/tex](3f(x)-4g(x))dx
=3[tex]\int\limits_{0}^{2}[/tex]f(x)dx-4[tex]\int\limits_{0}^{2}[/tex]g(x)dx
=3(-3)-4(2)
=-9-8
=-17

93. [tex]\int\limits_{2}^{4}[/tex](4f(x)-3g(x))dx
=[tex]\int\limits_{2}^{0}[/tex](4f(x)-3g(x))dx+[tex]\int\limits_{0}^{4}[/tex](4f(x)-3g(x))dx
=-[tex]\int\limits_{0}^{2}[/tex](4f(x)-3g(x))dx+[tex]\int\limits_{0}^{4}[/tex](4f(x)-3g(x))dx
=-(4(-3)-3(2))+(4(5)-3(-1))
=-(-12-6)+20+3
=41

If my answers are wrong, can someone please correct my mistakes?

[Math Tutor]

Nice work — 88, 88½, 90, 91, 92 and 93 are all correct.

The only slip is in 89, in the last step:
[tex]\int_2^4 (f(x)+g(x))\,dx = -(-3+2)+(5-1) = -(-1)+4 = 1+4 = 5[/tex]

Since [tex]-(-1)=+1[/tex], the answer is [tex]5[/tex], not [tex]3[/tex].