1.

Find \(\int_1^2\sqrt{x}-3x \,dx\).(a) \(\frac{8\sqrt{2}-31}{6}\)(b) \(8\sqrt{2}-31\)(c) \(\frac{\sqrt{2}-31}{3}\)(d) \(\frac{8\sqrt{2}+31}{4}\)This question was posed to me in an interview for job.My query is from Fundamental Theorem of Calculus-2 topic in section Integrals of Mathematics – Class 12

Answer»

Right ANSWER is (a) \(\frac{8\sqrt{2}-31}{6}\)

The best I can EXPLAIN: LET \(I=\int_1^2 \sqrt{x}-3x \,dx\)

F(x)=\(\int \sqrt{x}-3x \,dx\)

=\(\frac{x^{1/2+1}}{1/2+1}-\frac{3x^2}{2}=\frac{2x^{\frac{3}{2}}}{3}-\frac{3x^2}{2}\)

By using the second FUNDAMENTAL theorem of calculus, we get

I=F(2)-F(1)=\(\LEFT(\frac{2×2^{3/2}}{3}-\frac{3×2^2}{2}\right)-\left(\frac{2×1^{3/2}}{3}-\frac{3×1^2}{2}\right)\)

I=\(\frac{4\sqrt{2}}{3}-6-\frac{2}{3}+\frac{3}{2}=\frac{8\sqrt{2}-36-4+9}{6}=\frac{8\sqrt{2}-31}{6}\)


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