# Evaluate $\int_{0}^{1}2^{x^2+x}\mathrm dx$

Question : Evaluate - $$\int_{0}^{1}2^{x^2+x}\mathrm dx$$

My Attempt : First I tried to evaluate the indefinite integral of $2^{x^2+x}$ in order to put the limits $0$ and $1$ later on, but couldn't integrate it. Then I checked on WA and came to know that it's elementary integral doesn't exist. Now I moved one to using properties of definite integration such as $$\int_a^b f(x) \mathrm dx=\int_a^b f(a+b-x) \mathrm dx$$

But it couldn't help either. Can you please give me hint to proceed on this question?

P.S. - This is a high school level problem and therefore its solution shouldn't involve any special functions, such as Gaussian Integral etc.

Edit : I asked my teacher this question and basically this was an approximation based question. This was a MCQ type question which has an option "None of the above" and it was the correct answer, since the other options were made in such a way that can be rejected by bounding this integral between 2 functions. For example we can use $$2^{x^2+x}<2^{2x} ~; ~x\in (0,1)$$ and thus can be sure that this integral is less than $3/\ln(4)$.

Thanks all for devoting your time in my question!

• Looks to me like the gaussian integral you should be inspired by that integral to evaluate yours.. en.wikipedia.org/wiki/Gaussian_integral – Isham Sep 2 '17 at 4:19
• @Isham But this a high school level problem, and hence shouldn't involve Integrals like the one you mentioned. – Jaideep Khare Sep 2 '17 at 4:22
• $2^{x^2+x}$ is a complicated function. – Lord Shark the Unknown Sep 2 '17 at 4:24
• @Khare It looks the same for me...write $2 \int e^{(x+1)^2 ln 2 }$ you will see the gaussian integral more clearly...these kind of integrals arent simple at all... – Isham Sep 2 '17 at 4:25
• Inevitably, there is a link between this question and the Gaussian integral, which is easy to make out. However, I don't think the Gaussian integral, or it's evaluation falls under high school syllabus, although I knew about it. – астон вілла олоф мэллбэрг Sep 2 '17 at 4:27

If you cannot use special functions, then either numerical integration or approximation would be required.

For example, consider the Taylor expansion built around $x=\frac 12$ (mid point of the integration interval selected in order to tvoid promoting one of the bounds). You would get $$2^{x^2+x}=2^{3/4}+2^{3/4} \left(x-\frac{1}{2}\right) \log (4)+2^{3/4} \log (2) (1+\log (4))\left(x-\frac{1}{2}\right)^2+O\left(\left(x-\frac{1}{2}\right)^3\right)$$ Integrate termwise to get $$\int 2^{x^2+x}\,dx=2^{3/4} \left(x-\frac{1}{2}\right)+\frac{\left(x-\frac{1}{2}\right)^2 \log (4)}{\sqrt[4]{2}}+\frac{1}{3} 2^{3/4} \left(x-\frac{1}{2}\right)^3 \log (2) (1+\log (4))+O\left(\left(x-\frac{1}{2}\right)^4\right)$$ USe the bounds to get, as an approximation, $$\int_0^1 2^{x^2+x}\,dx\approx\frac{24+\log ^2(4)+\log (4)}{12 \sqrt[4]{2}}\approx 1.91361$$ while Wolfram Alpha would give $\approx 1.93749$.

For sure, you could improve using more terms. For illustration purposes, suppose that we make the expansion to $O\left(\left(x-\frac{1}{2}\right)^n\right)$. We should get $$\left( \begin{array}{cc} n & \text{result} \\ 2 & 1.91361 \\ 4 & 1.93589 \\ 6 & 1.93741 \\ 8 & 1.93749 \end{array} \right)$$

Hint:

Put $y = 2^{x^2+x}$

Integral becomes $\int_{1}^{4} \frac{1}{\sqrt{1+ \frac{4}{ln2}lny}}dy$

Again Put $\sqrt{1+ \frac{4}{ln2}lny} = u$

Integral becomes $\int_{1}^{3}\frac{1}{2e^a} e^{au^2}$

where $a = \frac{ln2}{4}$

It resembles the standard integral $\int_{1}^{3} e^{au^2}$

$$\int e^{au^2} = \frac{-i\sqrt{\pi}}{2\sqrt{a}} erf(iu\sqrt{a})$$

I hope you can take it from there

I am attaching the table of standard integrals for your reference

• I asked my teacher about this, and basically this was an approximation based question. This was a MCQ type question which has an option "None of the above" and it was the correct answer, since the other options were made in such a way that can be rejected by bounding this integral between 2 functions. For example we can use $$2^{x^2+x}<2^{2x} ~; ~x\in (0,1)$$ and thus can be sure that this integral is less than $3/\ln(4)$. – Jaideep Khare Sep 2 '17 at 8:49
Using $u=\frac{2x+1}{2}$ yields an imaginary error function.