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Whats the value of y ?


\displaystyle{ \qquad \sf \: y = \sf \int _0 }^(1) \sf (x {}^(2) + 5)\: \: dx


User LukeDuff
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2 Answers

3 votes
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Whats the value of y ? \displaystyle{ \qquad \sf \: y = \sf \int _0 }^(1) \sf (x {}^(2) + 5)\: \: dx-example-1
User Naveen Yedugani
by
5.0k points
5 votes

Answer:

5 1/3

Explanation:

You want the value of the definite integral of (x² +5) on the interval [0, 1].

Integral

The power rule applies:

∫(x^n)dx = (x^(n+1))/(n+1)

This means the integral is ...


\displaystyle \int_0^1{(x^2+5)}\,dx=\left((x^3)/(3)+5x\right)_0^1=(1)/(3)+5

The value of y is 5 1/3.

Whats the value of y ? \displaystyle{ \qquad \sf \: y = \sf \int _0 }^(1) \sf (x {}^(2) + 5)\: \: dx-example-1
User SMAG
by
5.0k points
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