If the value of ∫13(3x+5)dx+∫2b(2x−3)dx=28 with b>0, then the value of b=…
Explanation
We will calculate the value of each integral first.
Calculating the first integral
∫13(3x+5)dx=[23x2+5x]13
=(23(3)2+5(3))−(23(1)2+5(1))
=(227+15)−(23+5)
=227+230−23−210
=244=22
Calculating the second integral
∫2b(2x−3)dx=[x2−3x]2b
=(b2−3b)−(22−3(2))
=(b2−3b)−(4−6)
=b2−3b−(−2)
=b2−3b+2
Solving the equation
It is given that the sum of the two integrals is 28.
22+(b2−3b+2)=28
b2−3b+24=28
b2−3b−4=0
Factor the quadratic equation:
(b−4)(b+1)=0
This gives b=4 or b=−1.
Since the condition is b>0, the valid value is b=4.