How do you find #\int _ { - 2} ^ { 0} ( 2x + 5) d x#?

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Answer 1 · 2017-03-02T23:56:04

6 units squared.

Using the chain rule, the integral is #int_-2^0 x^2 + 5x dx#

=#x^2+5x]_-2^0# So we have to evaluate at the upper limit and evaluate the bottom limit and subtract them.

#(0)^2+5*(0)# = upper limit = 0

#(-2)^2+5*(-2)# = bottom limit = -6

Thus the answer is #0-(-6)# = 6 units squared.

Answer 2 · 2017-03-03T00:21:32

# int_(-2)^0 \ 2x + 5 \ dx = 6#

graph{2x+5 [-9.58, 10.42, -2.52, 7.48]}

The integrand is #y=2x+5#,and the integral represents the area under the curve from #x=2# to #x=0#

With # y=2x+5#:

# x=0 \ \ \ \ \=> y=5 #
# x=-2 => y=1 #

The area is a trapezium with #w=2, #a=1#, #b=5#

Hence,

# int_(-2)^0 \ 2x + 5 \ dx = 1/2(5+1)(2) = 6#