Question #98d56

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Answer 1 · 2017-04-07T05:57:34

graph{(x+5)/5 [-10, 10, -5, 5]}

#"Reqd. Area="int_1^5(x+5)/5 dx#

#=1/5[x^2/2+5x]_1^5#

#=1/5[{25/2+25}-{1/2+5}]#

#=1/5{25/2-1/2+25-5)#

#1/5(32)#

#=6.4" sq.unit."#

Otherwise, observe that the reqd. Area is the Area of a Trapezium

having parallel sides of length #6/5 (at x=1) and 2 (at x=5)# and the

dist. btwn. these parallel lines is #4=5-1.#

Hence, the reqd. area#=1/2(4){6/5+2)=2(16/5)=32/2=6.4" sq. unit"#