Evaluate the integral. ∫01(4x−9x2)dx

Question

Evaluate the integral.  ∫01(4x−9x2)dx

Ralph Lester · Accepted Answer

Step 1  Solve the integral ∫(4x−9x2)dx by using the general power rule of the integration that is ∫xn=xn+1n+1dx, where n is not equal to 1.  ∫(4x−9x2)dx=4x1+11+1−9x2+12+1  =4x22−9x33  =2x2−3x3  Step 2  Now, use the fundamental theorem of calculus and apply the upper and lower limit 1 and 0 in the integral solution of the integral ∫01(4x−9x2)dx to calculate the final value.  ∫01(4x−9x2)dx=2x2−3x3∣01  =2(1)2−3(1)3−[2(0)2−3(0)3]  =2-3-0  =-1

George Morin · Accepted Answer

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:

\int_{a}^{b} f (x) d x = F (x) ∣_{a}^{b} = F (b) - F (a)

Step 2: In this case,

f (x) = (4 x - 9 x^{2})

. Find its integral.

2 x^{2} - 3 x^{3} ∣_{0}^{1}

Step 3: Since

F (x) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(1)−F(0):

(2 \times 1^{2} - 3 \times 1^{3}) - (2 \times 0^{2} - 3 \times 0^{3})

Step 4: Simplify.
-1

Evaluate the integral. \int_{0}^{1}(4x-9x^{2})dx

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