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Calculus

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Evaluate the definite integral of the transcendental function.
the integral from 0 to 5 of (2^x +8)dx.
I got 40+ 31/log(2)but that's wrong.

  • Calculus - ,

    ∫2^x + 8 dx = 1/ln2 2^x + 8x
    so, using the limits, we have

    (2^5/ln2 + 40) - (2^0/ln2 + 0)
    = 32/ln2 + 40 - 1/ln2
    = 31/ln2 + 40

    you are correct.

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