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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.9
Evaluate the definite integrals in Exercises 176 to 195.
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Question : 178 of 271
Marks: +1, -0
∫12(4x2−5x2+6x+9)dx\int\limits_{1}^{2}(4x^2-5x^2+6x+9)dx
Solution:  
∫12(4x2−5x2+6x+9)dx\int\limits_{1}^{2}(4x^2-5x^2+6x+9)dx
=[4⋅x44−5⋅x33+6⋅x22+9x]12=\left[4\cdot\frac{x^4}{4}-5\cdot\frac{x^3}{3}+6\cdot\frac{x^2}{2}+9x\right]_1^2
=[x4−53x3+3x2+9x]12=\left[x^4-\frac{5}{3}x^3+3x^2+9x\right]_1^2
=(24−14)−53(23−13)+3(22−12)+9(2−1)=(2^4-1^4)-\frac{5}{3}(2^3-1^3)+3(2^2-1^2)+9(2-1)
=(16−1)−53(8−1)+3(4−1)+9(1)=(16-1)-\frac{5}{3}(8-1)+3(4-1)+9(1)
=643=\frac{64}{3}
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